51 Fully Solved Hard Problems in Quadratic Equations
Every problem below is worked in four stages — the question, two levels of hints, and a full solution — so you get stuck productively before you ever see an answer. Start with the reference sheet below, then work the 51 problems in any order.
Completing the square. For ax2+bx+c with a=0,
ax2+bx+c=a(x+2ab)2+(c−4ab2).
Every other identity below falls out of this single rewriting.
The quadratic formula. Setting ax2+bx+c=0 and solving the completed square for x gives
x=2a−b±b2−4ac.
The discriminant. Write D=b2−4ac. The quantity under the square root above governs everything about the nature of the roots:
D>0: two distinct real roots.
D=0: one repeated real root, x=−b/(2a).
D<0: two complex conjugate roots, no real root.
Sum and product of roots. If α,β are the roots of ax2+bx+c=0,
α+β=−ab,αβ=ac.
Conversely, the quadratic with a specified sum S and product P as its roots is x2−Sx+P=0.
Sign of a quadratic relative to its roots. For a>0 (the case a<0 is the mirror image), f(x)=ax2+bx+c is negative exactly between its two real roots (when D>0) and positive everywhere else.
xαβf(x)<0f(x)>0f(x)>0
For a>0, f(x)=ax2+bx+c dips below the axis only between its roots α,β.
Standard symmetric-function conversions. With α+β=−b/a and αβ=c/a written as S,P for brevity:
α2+β2=S2−2P,(α−β)2=S2−4P,α3+β3=S3−3PS.
How to use this page
Struggle first. Peek second.
Try the problem cold. If you're stuck after a genuine attempt, open Hint 1 — a small nudge, not the method. Still stuck? Hint 2 is a worked-solution skeleton with blanks for you to fill in. Only then check the full solution.
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Problem 51
Find the greatest negative integer satisfying x2−2x−99<0 and x2+x−20>0.
Understand the problem
We need to find a negative integer --- like −1, −2, or −3 --- that satisfies two conditions at once: x2−2x−99 must come out less than zero, and x2+x−20 must come out greater than zero. Several negative integers will satisfy both conditions --- we want the one closest to zero (for example, −2 is greater than −7, since it's closer to zero).
Four conditions on a real number x are given. The first condition is a quadratic inequality, giving a validity range for x. The second condition is another quadratic inequality, giving a second validity range for x.
The third condition is hidden in the wording: x must be negative. The fourth condition is hidden too: x must be an integer.
The question asks for the x satisfying all four conditions at once.
The strategy is to intersect the four ranges these conditions describe.
Condition 1: the first inequality factors as
x2−2x−99<0⇒(x−11)(x+9)<0⇒−9<x<11.
Condition 2: the second inequality factors as
x2+x−20>0⇒(x−4)(x+5)>0⇒x<−5 or x>4.
Condition 3: the question asks for a negative x, so
x<0.
Condition 4: the question asks for an integer x, so
x∈Z.
All four conditions are AND conditions --- they must hold simultaneously. Intersecting Conditions 1 and 2 gives −9<x<−5 or 4<x<11; Condition 3 discards the second branch, leaving
−9<x<−5.
Condition 4 restricts this further to integers.
The greatest integer satisfying −9<x<−5 is xmax=−6.
A common silly mistake is to answer −5: it looks like the boundary of the interval, but the inequality is strict, so −5 itself is excluded.
Take the 2026 roots of unity. Subtract each of them from 1, discard the zero, and multiply the rest. Compute the result.
Understand the problem
The `2026 roots of unity' are the 2026 numbers that equal 1 when raised to the power 2026 --- one of them is 1 itself, and the rest are spread evenly around a circle of complex numbers. We need to take each of these 2026 numbers, subtract it from 1, and throw away the one result that comes out to zero (that's the 1−1 case), then multiply everything else together to get a single number.
The nnth roots of unity are the complex numbers under discussion.
The question asks for the product of every root minus 1, except the root equal to 1 itself. Keep that exclusion in mind -- it's exactly what turns up again later in the strategy.
The strategy is to write the defining polynomial as a product of one linear factor per root, then divide out the one factor you already know. What's left holds for every number, not just the roots -- so plug in the number that turns the product you want into a simple sum on the other side.
Label the nnth roots of unity as 1,ζ1,ζ2,…,ζn−1, where 1 is real and the rest are complex. These are exactly the roots of the equation xn−1=0, so
xn−1=(x−1)(x−ζ1)(x−ζ2)⋯(x−ζn−1).
Divide both sides by (x−1):
1+x+x2+⋯+xn−1=(x−ζ1)(x−ζ2)⋯(x−ζn−1).
This identity holds for every x, not just the roots. Substitute x=1.
Investigate the range of f(x) if x is real and:
f(x)=tan(x−192π)tan(x+192π).
Understand the problem
As x takes every possible real-number value, f(x) produces a whole set of outputs --- we need to describe exactly which numbers are in that set (that's what range means). Nothing needs to be solved for one particular x; we're describing the full spread of possible output values at once.
The expression is a ratio of two tangents: x plus a fixed angle, over x minus that same angle. x ranges over all reals where both tangents are defined.
As x varies, this ratio traces out many values. The question asks for the values it never reaches.
Testing values of x directly won't reveal this, since tangent is periodic and unbounded. The ratio can instead be rewritten to depend on x through a single bounded quantity, sin2x.
The strategy is to solve for sin2x in terms of the ratio, then use that ∣sin2x∣≤1 to turn this into an inequality in the ratio itself. Factoring that inequality as a difference of squares pins down the two values the ratio can never reach -- and simplifying those two values with a half-angle identity gives the answer.
Write φ=192π. Then
y=tan(x−φ)tan(x+φ)=cos(x+φ)sin(x−φ)sin(x+φ)cos(x−φ).
Using the product-to-sum identities sinAcosB=21[sin(A+B)+sin(A−B)] and cosAsinB=21[sin(A+B)−sin(A−B)] with A=x+φ,B=x−φ:
y=sin2x−sin2φsin2x+sin2φ.
Let s=sin2x∈[−1,1]. Solving for s:
y(s−sin2φ)=s+sin2φ⟹s(y−1)=sin2φ(y+1)⟹s=y−1sin2φ(y+1)(y=1).
Since ∣s∣≤1:
sin2(2φ)(y+1)2≤(y−1)2⇒(y−1)2−sin2(2φ)(y+1)2≥0.
Factor as a difference of squares, with c=sin2φ:
[(y−1)−c(y+1)][(y−1)+c(y+1)]⟹[y(1−c)−(1+c)][y(1+c)−(1−c)]≥0≥0.
This product of two linear factors in y is non-negative outside the interval between their roots, y=1+c1−c and y=1−c1+c, as Figure~\figref{D3forbidden} shows. So
y∈/(1+sin2φ1−sin2φ,1−sin2φ1+sin2φ).
y1+c1−c1−c1+cy forbidden hereallowedallowed
y is forbidden strictly between the two roots, free everywhere else.
Note that
1+sin2φ1−sin2φ=(cosφ+sinφ)2(cosφ−sinφ)2=tan2(4π−φ).
The same steps, with the two square roots swapped, give
1−sin2φ1+sin2φ=(cosφ−sinφ)2(cosφ+sinφ)2=tan2(4π+φ).
So y takes every real value except those strictly between the two bounds:
y∈(−∞,tan2(4π−192π)]∪[tan2(4π+192π),∞).
The real numbers x1,x2,x3 satisfying x3−x2+βx+γ=0 are in geometric progression. Show that γ=−β3, and find the values β can take.
Understand the problem
`Geometric progression' means each number is the one before it times a fixed ratio. The equation's three roots x1,x2,x3 form such a sequence. We first need to show γ=−β3 follows from this, then find every value β can take.
Three real numbers are the roots of a monic cubic with two free coefficients. These three numbers must form a geometric progression.
The question asks you to show the two coefficients satisfy one identity forced by the GP condition, then find every value one of them can take.
Tip: you don't need to name all three GP terms with a starting term and a ratio. The product of three terms in geometric progression is always just the middle term cubed --- so only the middle term needs a name.
Call the middle term r. Since r is itself one of the three roots, it must satisfy the cubic --- substitute it in, simplify, and you'll find β and r are actually the same number, which is exactly what lets you pin down γ in terms of β.
Knowing that β itself is a root lets you divide it out of the cubic, leaving a quadratic that the other two roots satisfy. Demand that quadratic's discriminant be positive, so those two roots come out real and distinct --- that gives you a range for β. Then pause: is there some value of β inside that range where the roots secretly fail to be distinct anyway? Check it, and exclude it if so.
Let the GP roots be ρr,r,rρ. Using the coefficient-root relations for x3−x2+βx+γ=0, write their product in terms of γ: product =
r is itself a root, so it satisfies the cubic directly. Substitute your expression for the product into the cubic and simplify, using r=0, to solve for β in terms of r, then for γ in terms of β:
β=,γ=
You now know one root is β itself. Divide the cubic by (x−β) to find the quadratic satisfied by the other two roots:
x3−x2+βx−β3=(x−β)
For the other two roots to be real and distinct, demand that this quadratic's discriminant be strictly positive:
>0⇒β∈
Is there some value of β inside this range where the quadratic's roots secretly coincide with r=β itself, breaking distinctness? Check the quadratic at β=0: its roots become .
Let the roots in geometric progression be ρr,r,rρ for some ratio ρ=0,±1. Their product is r3, which by the coefficient-root relations equals −γ:
r3=−γ.
Since r is itself one of the three roots, it satisfies the cubic directly:
r3−r2+βr+γ=0.
With γ=−β3 and the known root r=β, divide the cubic by (x−β):
x3−x2+βx−β3=(x−β)(x2+(β−1)x+β2).
For the remaining two roots to be real and distinct, this quadratic's discriminant must be strictly positive:
(β−1)2−4β2>0⇒−3β2−2β+1>0⇒−(β+1)(3β−1)>0⇒β∈(−1,31).
At β=0, the quotient becomes x2−x=x(x−1), giving roots 0,1. But x=0 coincides with the already-found root r=β=0, so the cubic's actual root set is {0,0,1}, not three distinct values. So β=0 must be excluded.
Combining this with the discriminant condition:
β∈(−1,31)∖{0}.
In words: β can be any number strictly between −1 and 31, except 0.
A function f:R→R is defined by: f(x)=α+6x−8x2αx2+6x−8. Find the values of α for which f is onto, and show that f is one-to-one for none of them.
Understand the problem
`Onto' means every output hits some input; `one-to-one' means no two inputs share an output. We first need to find every α making f onto, then show none of those α make f one-to-one.
f(x) is a ratio of two quadratics in x, and both quadratics are built from the same unknown parameter α.
The question first asks for every α that makes f's range all of R --- in other words, makes f onto. Then, restricted to just those α, it asks whether any of them also make f one-to-one.
These aren't really two separate questions --- the same underlying quadratic answers both.
Set f(x) equal to an arbitrary target y and cross-multiply. This turns the equation into a quadratic in x, with α and y both appearing in its coefficients. For f to be onto, every real y must come from some real x --- so that quadratic's discriminant, which is itself a quadratic in y, must never go negative: it must open upward, and never dip below the axis. Demanding both of those conditions pins down exactly which α make f onto.
For those α, ask how many x map to the same y. Away from the very edge of the range, the discriminant stays strictly positive for every y, so every output has two different inputs landing on it --- so f is never one-to-one. Right at the edge, only one particular y has just a single input, but every other y still has two --- so f still isn't one-to-one there either.
f is onto exactly on a closed interval of α-values.
Set f(x)=y, cross-multiply: x2()+x()+()=0
Demand a real x for every real y: discriminant (in x) ≥0 for all y.
Discriminant =Disc(y), a quadratic in y with leading coefficient
For Disc(y)≥0 for all y: leading coefficient must be positive, and Disc(y)'s own discriminant Δy≤0.
Δy=⋅(α−)(α−)
Combine both conditions.
Onto exactly when α∈
One-to-one means each y has at most one x-preimage. Check the sign of Disc(y) strictly inside this range.
For α strictly inside this interval, Disc(y)≥≤0 for every y, so every y has \rule[-0.2em]{0.65cm}{1.5pt} x-preimages
Now check the two endpoints of the range: what does Δy equal there, and how does Disc(y) behave at one particular y versus every other y?
At the endpoints, Disc(y)=0 for \rule[-0.2em]{0.65cm}{1.5pt} value(s) of y, and Disc(y)>0 for \rule[-0.2em]{0.65cm}{1.5pt}
Does that one special y stop f from being onto? Does every other y still have two preimages?
Hence, f is one-to-one for values of α in this range
Setting f(x)=y and cross-multiplying:
y(α+6x−8x2)=αx2+6x−8⟹x2(−8y−α)+x⋅6(y−1)+(αy+8)=0.
For a real x to exist for a given y, the discriminant (in x) must be non-negative:
Disc(y)=36(y−1)2+4(8y+α)(αy+8)=(36+32α)y2+(184+4α2)y+(36+32α).
For f to be onto, Disc(y) --- itself a quadratic in y --- must be non-negative for every y. Its leading coefficient 36+32α must therefore be positive, so that Disc(y) opens upward. An upward-opening quadratic dips below zero exactly when it has two distinct real roots, i.e.\ when its own discriminant is positive (Figure~\figref{D5disc}). So to stay non-negative everywhere, Disc(y)'s own discriminant Δy must satisfy Δy≤0.
Computing Δy as a difference of squares:
Δy=(184+4α2)2−4(36+32α)2=[(184+4α2)−2(36+32α)][(184+4α2)+2(36+32α)]=(4α2−64α+112)(4α2+64α+256)=16(α2−16α+28)(α2+16α+64)=16(α−2)(α−14)(α+8)2.
Since (α+8)2≥0 always, Δy≤0 exactly when (α−2)(α−14)≤0, i.e.\ α∈[2,14], and this entire interval satisfies 36+32α>0 automatically. So
f is onto exactly when α∈[2,14].
yDisc(y)≥0 for every y
The curve stays ≥0 everywhere, touching the axis at most once.
α214f is onto
f is onto exactly on the closed interval [2,14].
For the one-to-one question: at any α strictly inside (2,14), Δy<0, so Disc(y)>0 for every real y --- every output y has two distinct real x-preimages.
At the endpoints α=2 or α=14, Δy=0. This means Disc(y) touches zero at exactly one particular value of y, and stays strictly positive at every other value of y.
That one special y still has a real x with f(x)=y --- it just has only one such x instead of two. So f still reaches every y, and is still onto.
But every other y has Disc(y)>0, giving two distinct x-preimages. That alone is enough to show f is not one-to-one at α=2 or α=14 either.
In every case within [2,14], f fails to be one-to-one.
Let f be a quadratic that is positive for every real x. Pick real numbers m,k with 2k≥m2, and form g=f+mf′+kf′′. Show that g is positive for every real x too.
Understand the problem
f′ and f′′ are f's first and second derivatives, measuring how f changes and how that change itself changes. We combine f with both, weighted by m,k, to build a new function g. We need to show g stays positive everywhere too.
f is a quadratic that stays strictly positive everywhere. A new function g is built by adding a multiple of f′ and a multiple of f′′ to f itself. The two multipliers satisfy an inequality. Twice the second is at least the square of the first.
The question asks you to confirm g inherits the same property: strictly positive for every real x.
Testing specific numbers won't work. The claim must hold for every f and every admissible pair of multipliers at once. What matters is that g is still a quadratic, so the same positivity test that applies to f applies to g directly.
Write f in the standard form that makes "positive everywhere" easy to test: positive leading coefficient, negative discriminant. Differentiate to get f′ and f′′, substitute into g, and collect terms --- g turns out to have the same leading coefficient as f. Compute g's discriminant in terms of f's; the given inequality on m,k is exactly what's needed to show it stays negative too, which is enough to conclude g is positive everywhere.
Write f(x)=ax2+bx+c with a>0 and discriminant D=b2−4ac<0 --- the standard characterization of a quadratic that is positive for every real x.
Then f′(x)=2ax+b and f′′(x)=2a, so
g(x)=f(x)+mf′(x)+kf′′(x)=ax2+(b+2am)x+(c+bm+2ak).
The leading coefficient of g is still a>0. Its discriminant is
(b+2am)2−4a(c+bm+2ak)=b2+4abm+4a2m2−4ac−4abm−8a2k=D+4a2(m2−2k).
Since D<0 and 2k≥m2 gives m2−2k≤0, this discriminant is strictly negative:
D+4a2(m2−2k)≤D<0.
A quadratic with positive leading coefficient and negative discriminant is positive for every real x (Figure~\figref{D6pos}). Hence g(x)>0 for all real x.
Three quadratics share the same squared leading coefficient. f and g have opposite signs on their other two coefficients. h doubles those same two coefficients, keeping f's sign pattern. α is a root of f, β a root of g, both positive, with α the smaller.
The question asks you to show h(α)h(β)<0.
α, β, and the roots of h have no exact formula --- the coefficients are arbitrary. What matters is not their value, but the sign h takes at these two points.
Since α satisfies f(α)=0, solve that equation for the shared constant term; do the same with β in g(β)=0. Substituting each of these back into h collapses most of the expression, leaving just a multiple of α2 in one case and of β2 in the other --- with opposite signs, since f and g's coefficients were opposite to begin with. Since α,β>0, those signs carry straight through to h(α) and h(β).
Let n≥2, and suppose n quadratic equations x2+p1x+q1=0,…,x2+pnx+qn=0 are arranged in a circle, so that each equation shares a root with each of its two neighbors. Prove that: p12+⋯+pn2≥2(q1+⋯+qn).
Understand the problem
Picture n quadratics in a circle, each sharing a root with its neighbors. We need to prove the sum of every pk2 is always at least twice the sum of every qk.
n quadratic equations are arranged in a circle, each sharing a root with both of its neighbors. This gives n shared roots in total, one per pair of neighboring equations.
The question asks you to show the sum of the squares of all the middle coefficients is always at least twice the sum of all the constant terms, and to pin down exactly when the two sides are equal.
Name the root shared by equation k and equation k+1 as rk. Then equation k's own two roots are just rk−1 and rk, so its coefficients follow directly from those two roots by the usual sum/product relations. Square and sum that sum-relation across the whole circle: it splits into twice a sum of squares of the rk's, plus twice a sum of products rk−1rk --- and that second piece is exactly the sum of all the constant terms. Rearranging gives the inequality immediately, since a sum of squares can never be negative --- and equality holds exactly when every shared root is 0.
Name the root shared by equation k and equation k+1 (indices mod n) as rk, so equation k's own two roots are rk−1 and rk, giving
pk=−(rk−1+rk),qk=rk−1rk.
Square and sum:
k=1∑npk2=k=1∑n(rk−1+rk)2=2k=1∑nrk2+2k=1∑nrk−1rk=2k=1∑nrk2+2k=1∑nqk.
So
k=1∑npk2−2k=1∑nqk=2k=1∑nrk2≥0,
i.e.\ p12+⋯+pn2≥2(q1+⋯+qn), for every n≥2. This uses only the defining relations directly, never solving for the rk individually, so it needs no assumption on n's parity.
Equality holds exactly when ∑krk2=0, i.e.\ when every rk=0: every pairwise shared root is 0.
Let α,β be the roots of the equation ax2+bx+c=0. Find the equation whose roots are (α2−β2)(α3−β3) and α3β2+α2β3.
Understand the problem
Starting from one quadratic with roots α,β, we build two new numbers by combining specific powers of α and β. We need to find the quadratic equation whose two roots are exactly these two numbers.
α,β are known only through their sum and product, via the coefficient-root relations on ax2+bx+c=0.
The question asks for the quadratic equation with roots (α2−β2)(α3−β3) and α3β2+α2β3.
Both of these expressions can be factored down to a difference α−β times something symmetric, or a power of αβ times something symmetric --- and everything symmetric in α,β reduces to their sum and product. So neither α,β individually, nor α−β itself, ever needs to appear in the final answer.
Factor α2−β2 and α3−β3 each as (α−β) times a symmetric piece, and multiply --- the two (α−β) factors combine into (α−β)2, which is symmetric too. For the second expression, pull out the common factor α2β2, leaving just α+β behind. Once both target expressions are written purely in terms of the sum and product, their own sum and product give the coefficients of the equation you want.
Investigate the range of x2+qx+1x2+px+1, where p,q are real numbers with ∣p∣<2 and ∣q∣<2.
Understand the problem
As x ranges over every real number, the fraction x2+qx+1x2+px+1 produces a whole set of possible outputs --- and we need to describe that entire set (its range). The numbers p,q are fixed in advance, each somewhere strictly between −2 and 2.
A ratio x2+qx+1x2+px+1 is given, with −2<p,q<2. This range keeps both numerator and denominator positive for every real x. The variable x ranges over all reals.
As x varies, this ratio takes a range of values. The question asks exactly which values are achievable.
A value is achievable exactly when some real x produces it. This turns the question into finding which target values make an equation in x have real solutions.
Set the ratio equal to a target value y and cross-multiply --- this gives a quadratic in x whose coefficients involve y. Demand that quadratic's discriminant be non-negative, and factor the resulting inequality as a difference of squares in y. This pins down two critical values of y, using that −2<p,q<2 along the way; the achievable values of y turn out to be everything between them, endpoints included, since the discriminant hits exactly 0 there.
Let y=x2+qx+1x2+px+1. Since ∣q∣<2, the denominator x2+qx+1 has discriminant q2−4<0, so it is never zero. The ratio is defined for every real x.
Cross-multiplying and collecting as a quadratic in x:
(y−1)x2+(yq−p)x+(y−1)=0.
For a real x to exist, the discriminant must be non-negative:
(yq−p)2−4(y−1)2≥0.
Factoring as a difference of squares:
[(yq−p)−2(y−1)][(yq−p)+2(y−1)]≥0⟹[(2−p)−(2−q)y][(2+q)y−(2+p)]≥0.
Since ∣p∣<2 and ∣q∣<2, both 2−q>0 and 2+q>0. This product of two linear factors in y vanishes at
y=2−q2−pandy=2+q2+p,
and is negative as y→±∞, so it is non-negative exactly between these two roots.
Hence, as Figure~\figref{D10range} shows,
y∈[min(2−q2−p,2+q2+p),max(2−q2−p,2+q2+p)].
ymin(2−q2−p,2+q2+p)max(2−q2−p,2+q2+p)attainable y
Let B,C be real numbers. Find D and E such that x4+Bx3+Cx2+Dx+E=0 is the square of a quadratic, and find those two roots in terms of B,C.
Understand the problem
`The square of a quadratic' means the equation factors as the square of some quadratic --- so there are really only two solutions, each repeated. B,C are fixed; we need to find D,E, and then those two repeated solutions.
The quartic x4+Bx3+Cx2+Dx+E=0 is given, with B,C free real parameters and D,E left to be determined.
The question asks for D,E in terms of B,C so the quartic is a perfect square of some quadratic. That means it has just two distinct roots, each repeated twice. It also asks for those two repeated roots.
There is no need to factor the quartic or hunt for roots first. ``Equals the square of a quadratic'' is a statement about the quartic's coefficients directly, found by expanding a general squared quadratic and matching term by term. Once D,E are pinned down this way, the quartic's two repeated roots are just the roots of that quadratic itself, found directly by the quadratic formula.
Suppose the roots of x2−ax+b=0 are real and they differ by a quantity that lies between d and c, with 0≤d<c. Find the range of values that b is permitted to take (as a function of a,c,d).
Understand the problem
`Differ by' just means the gap between the two roots --- the bigger one minus the smaller one. We're told that gap has to land strictly between two fixed numbers, d and c (with d smaller). We need to find which values b can take to make this happen, written in terms of the other letters a,c,d.
The quadratic x2−ax+b=0 has real roots α,β. Their difference is bounded: less than c, and more than d.
The question asks for the range of b satisfying both bounds.
Working with α,β individually introduces a square root. Instead, use that (α−β)2 has a clean closed form in a,b.
Express (α−β)2 via the sum and product of the roots. The upper bound on the difference gives one inequality on b. The lower bound gives another. Combine both.
Let α,β be the roots of ax2+2bx+c=0, and suppose mα+δ,mβ+δ are the roots of Ax2+2Bx+C=0. Investigate the constraint that this places on a,b,c,A,B,C.
Understand the problem
We take the two roots of the first equation and transform them the same way --- multiply each by m, then add δ --- to get two new numbers. We're told these new numbers turn out to be exactly the roots of a second equation, built from A,B,C. We need to find what relationship must hold among all six letters a,b,c,A,B,C for that to be possible.
ax2+2bx+c=0 has roots α,β. Ax2+2Bx+C=0 has roots mα+δ,mβ+δ, for some unspecified real m=0 and δ.
The question asks for the constraint relating a,b,c to A,B,C.
Finding m,δ explicitly introduces unknowns that will not fully cancel. Instead, use that shifting every root by δ leaves the difference α−β unchanged, while scaling by m scales that difference by m.
Express (α−β)2 for each quadratic in terms of its own coefficients. Relate the second equation's root-difference to the first's: the shift drops out, only m survives. Equate the two expressions and rearrange to isolate the ratio in terms of a,A,m alone.
Let α,β be the roots of ax2+2bx+c=0, and let mα+δ,mβ+δ be the roots of Ax2+2Bx+C=0, for some real m=0 and real δ.
For a quadratic px2+2qx+r=0 with roots u,v: u+v=−2q/p, uv=r/p, so
(u−v)2=(u+v)2−4uv=p24q2−p4r=p24(q2−pr).
Applying this to the first equation: (α−β)2=a24(b2−ac).
The shift δ cancels in any difference of the transformed roots, while the scale factor m carries through:
(mα+δ)−(mβ+δ)=m(α−β).
So [(mα+δ)−(mβ+δ)]2=m2(α−β)2.
Applying the same root-difference formula to the second equation:
[(mα+δ)−(mβ+δ)]2=A24(B2−AC).
Combining the last two displays:
A24(B2−AC)=m2(α−β)2=m2⋅a24(b2−ac).
Cancelling the common factor of 4 and rearranging:
B2−ACb2−ac=m2A2a2=(mAa)2.
Suppose α+β=s and α3+β3=t, for real numbers s and t, s=0. Find the quadratic equation whose roots are α and β.
Understand the problem
We're only told two facts about α and β --- their sum, and the sum of their cubes --- not what α and β actually equal. Using just these two clues, we need to reconstruct the quadratic equation that has α and β as its two roots.
The question asks for the quadratic equation with roots α,β, in terms of s,t.
Writing that equation needs α+β and αβ. The sum is already given. The product is not. Use the identity connecting a sum of cubes to the sum and product.
Expand α3+β3 using that identity. Substitute s,t. This leaves one equation in αβ alone, solved by dividing through by s (so s=0 is required). With sum and product in hand, the equation follows.
Suppose (x−3m)(x−m−3)<0. Find the range of m for which this holds (i) for every x∈[p,q], and (ii) for at least one x∈[p,q].
Understand the problem
The expression (x−3m)(x−m−3) comes out negative exactly when x sits strictly between its two roots. For example, if m=1 the two roots are 3 and 4; if m=2 they are 6 and 5 instead, with 6 now the larger one. We need to find two different ranges of m. First, the values that make this true for every x in the fixed interval [p,q]. Second, the weaker values that make it true for at least one x in [p,q].
(x−3m)(x−m−3)<0 is considered against the interval [p,q], in two senses.
The question asks for the values of m, in terms of p,q, so that (i) the inequality holds for every x∈[p,q], and (ii) it holds for at least one x∈[p,q].
For (i), matching the roots to the endpoints is not the same as requiring the inequality throughout: this is a containment relationship between intervals. For (ii), reusing that containment logic is wrong: holding somewhere is a weaker demand, met when the two intervals merely touch.
The product is negative exactly between the two roots 3m,m+3. Compare the two roots directly to see which one is smaller -- this splits into two cases depending on m. For (i), require [p,q] to sit strictly inside the root-interval. For (ii), require only that the two intervals overlap.
root-interval[p,q](i) containment: every x∈[p,q] worksroot-interval[p,q](ii) overlap: some x∈[p,q] works
Find the condition under which ax2+bx+c=0 has real roots that always lie outside (−k,k).
Understand the problem
`Outside (−k,k)' means the two roots straddle the interval: one root at or below −k, the other at or above k, so −k and k both sit strictly between the two roots. We need to find the exact condition on a,b,c,k that guarantees this.
f(x)=ax2+bx+c has real coefficients. k>0 is a bound.
The question asks for the condition on a,b,c,k exactly equivalent to f having real roots α,β with α<−k<k<β.
Both directions need justification. A condition derived only by assuming the roots straddle would just be necessary. A separate argument is needed to show it is also sufficient.
The coefficients are arbitrary, so the roots cannot be found explicitly. Use instead the sign of f at k and at −k, and, for the reverse direction, that a⋅f(x)→+∞ far enough from the origin in either direction.
Evaluate a⋅f(k) and a⋅f(−k) assuming the straddling roots exist, to derive a candidate inequality. For the reverse direction, start from that inequality and show it forces a⋅f negative at both k and −k. Since a⋅f→+∞ at both ends, it must cross zero somewhere beyond k and somewhere beyond −k.
Let f(x)=ax2+bx+c. We show the condition k2+ac+kab<0 is exactly equivalent to f having real roots α,β with α<−k<k<β (Figure~\figref{D16order}).
α−kkβboth k,−k inside (α,β)
Both k and −k lie strictly inside the roots' interval.
Necessity.
Suppose α<−k<k<β. Then k and −k both lie strictly between the roots, where f has sign opposite to a. So a⋅f(k)<0 and a⋅f(−k)<0:
a⋅f(k)=a2k2+abk+ac<0,a⋅f(−k)=a2k2−abk+ac<0.
Dividing each by a2>0:
k2+ac<−abk,k2+ac<abk.
Both must hold simultaneously, so k2+ac is less than whichever of ±abk is smaller, that is, less than −abk:
k2+ac<−abk⟹k2+ac+kab<0.
Sufficiency.
Suppose k2+ac+kab<0. Since −abk≤±abk for either sign, this gives
k2+ac<−abk≤−abk⟹k2+abk+ac<0⟹a⋅f(k)<0,
and likewise k2+ac<−abk≤abk gives a⋅f(−k)<0.
a⋅f(x)=a2x2+abx+ac→+∞ as x→+∞ (the a2x2 term dominates), but a⋅f(k)<0. Since a⋅f is continuous, it must cross zero somewhere beyond k -- a root of f strictly beyond k. Likewise, a⋅f(−k)<0 and a⋅f→+∞ as x→−∞ force a root strictly beyond −k in the negative direction.
A quadratic has at most two roots, so these are exactly α,β, giving α<−k<k<β.
Suppose x2+abx+c=0 and x2+acx+b=0 share a common root. Find the equation whose roots are the other two roots.
Understand the problem
Each of the two given equations has two roots. One of those roots is shared --- the exact same root solves both equations. We need to find the equation whose roots are those two unshared roots, along with the condition on a,b,c that makes a shared root possible.
The question asks for the equation whose roots are the two leftover roots, one from each equation. It also asks for the condition on a,b,c that makes a shared root possible at all.
Solving both quadratics explicitly is unnecessarily heavy. Instead, subtract the two equations directly.
Subtracting cancels the x2 term, leaving a linear equation that pins down the shared root directly. Use that root with the product-of-roots relation on each equation to find each leftover root. Their sum and product give the target equation. Separately, substitute the shared root back into either original equation, not just the difference, to get the genuine existence condition.
Use the product-of-roots relation on each original equation.
Other root of Eqn.~(1), from t⋅(other root)=c:
Other root of Eqn.~(2), from t⋅(other root)=b:
Form the equation from these two leftover roots.
Hence, the equation is x2−x+=0
A shared root forced by subtraction is only genuine if it satisfies an original equation --- check it.
Substitute t=1/a into Eqn.~(1) directly: a21+b+c=
Hence, a common root can exist at all only if b+c=
Let t be the common root of x2+abx+c=0 and x2+acx+b=0. Subtracting the two equations, the x2 terms cancel:
(ab−ac)t+(c−b)=0⟹a(b−c)t=b−c⟹t=a1(for b=c).
Using the product-of-roots relation on the first equation (t⋅q1=c, where q1 is its other root):
a1⋅q1⟹q1=c=ac.
Using the product-of-roots relation on the second equation (t⋅q2=b):
a1⋅q2⟹q2=b=ab.
The equation with roots q1=ac and q2=ab has sum ac+ab=a(b+c) and product ac⋅ab=a2bc:
x2−a(b+c)x+a2bc=0.
This derivation only shows that if a common root exists, it must equal 1/a. Subtracting the two equations cannot confirm this value genuinely satisfies either original equation, only that it is the unique candidate.
To check existence, substitute t=1/a into the first equation:
a21+ab⋅a1+c⟹b+c=a21+b+c=0=−a21.
So a common root exists only when b+c=−a21. For any a,b,c not satisfying this, the two equations share no root.
If one root of the equation ax2+bx+c=0 is the cube of the other, then prove that b4−4ab2c=ac(a−c)2.
Understand the problem
`One root is the cube of the other' means if one root is some number r, the other root is exactly r3 --- not just any related number, a very specific one (for example, if one root were 2, the other would have to be 23=8). We need to show this special relationship forces one exact equation connecting a,b,c to be true.
ax2+bx+c=0 has a,b,c all nonzero, and one root is the cube of the other: roots α,α3.
The question asks you to show this forces b4−4ab2c=ac(a−c)2.
There is no clean sum/product statement for a cube-root relation the way there is for, say, one root being twice the other. The two coefficient-root relations here, α+α3=−b/a and α⋅α3=c/a, both mix α and α3 together, with no clean way to isolate either alone.
Name β=α2. Squaring the sum relation turns it into a relation purely in β, and the product relation is already purely in β once you notice α⋅α3=α4=β2. Combining these two β-relations eliminates α entirely, pinning down β in terms of a,b,c. Substituting that β back into β2=c/a and clearing denominators reaches the identity directly.
For the equation x2−(k+p)x+(k2+k−q)=0, where p,q are fixed real numbers, suppose one root exceeds 2 and the other is less than 2. Find the set of values k can take, in terms of p and q.
Understand the problem
`One root exceeds 2 and the other is less than 2' means the number 2 sits strictly between the two roots. p and q are fixed numbers we don't get to choose. k is the one thing we're solving for. We need to find every value of k that makes this happen, written in terms of p and q.
f(x)=x2−(k+p)x+(k2+k−q) has one root above 2 and one below 2.
The question asks for the set of values of k, in terms of p,q.
Don't try to find the two roots explicitly --- you'll get tangled in a square root for no reason. Look instead at the sign of f at 2.
Evaluate f(2), and use that, since 2 lies strictly between the roots, this value must be opposite in sign to the leading coefficient. This gives a quadratic inequality in k, whose solution set is the answer.
Let f(x)=x2−(k+p)x+(k2+k−q). Since one root exceeds 2 and the other is less than 2, the value 2 lies strictly between the roots, where f has sign opposite to its leading coefficient (=1>0), as shown in Figure~\figref{D19root}. So f(2)<0:
f(2)=4−2(k+p)+(k2+k−q)=k2−k+(4−2p−q)<0.
By the quadratic formula, k2−k+(4−2p−q)=0 has real roots when 8p+4q−15≥0:
k=21±1−4(4−2p−q)=21±8p+4q−15.
If 8p+4q<15, the left side k2−k+(4−2p−q) is always positive, and no value of k satisfies the straddling condition.
When 8p+4q≥15, the upward-opening parabola in k is negative exactly between its two roots, so
k∈(21−8p+4q−15,21+8p+4q−15).
Solve [a2+2(1−b)]x2+2a(1+b)x+2b(b−1)+a2=0, if the equation has equal roots.
Understand the problem
`Equal roots' means the equation's two solutions are actually the same number --- a repeated root, not two different ones. Here a and b are treated as fixed, already-known numbers. We need to solve for x itself, under that special repeated-root condition.
[a2+2(1−b)]x2+2a(1+b)x+[2b(b−1)+a2]=0 has equal roots.
The question asks for the relation this forces between a,b, and then the common value of the repeated root.
Setting the discriminant to zero produces a messy fourth-degree expression in a,b. Notice a appears only through a2, so the expression is really a quadratic in u:=a2.
Write the discriminant condition as a quadratic in u, with b fixed. It factors into two pieces: the target relation, and a sum of two squares that only vanishes in a degenerate case. Discard the degenerate case. Substitute the relation into the repeated-root formula.
Equal roots requires the discriminant of [a2+2(1−b)]x2+2a(1+b)x+[2b(b−1)+a2]=0 to vanish:
[2a(1+b)]2−4[a2+2(1−b)][2b(b−1)+a2]=0.
Dividing by 4 and expanding, then collecting as a quadratic in u:=a2:
u2+u(b2−6b+1)−4b(b−1)2=0.
This factors as
(u−4b)(u+(b−1)2)=0.
Since u=a2≥0 and (b−1)2≥0, the second factor u+(b−1)2=0 only when a=0 and b=1 simultaneously.
But then the leading coefficient a2+2(1−b)=0, so the equation is not a genuine quadratic. Excluding this degenerate case, the first factor must vanish:
a2=4b.
For the repeated root, use x=−2⋅leading coefficientmiddle coefficient:
x=2[a2+2(1−b)]−2a(1+b)=a2+2−2b−a(1+b).
Substituting a2=4b into the denominator:
a2+2−2b=4b+2−2b=2(b+1).
So the root is
2(b+1)−a(1+b)=−2a.
This is valid for b=−1. Combined with a2=4b, b=−1 would need a2=−4, which is impossible for real a -- so this case never actually arises.
Discuss the nature of the roots of (a+2b−3c)x2+(b+2c−3a)x+(c+2a−3b)=0, given that a,b,c are rational. Find the condition on a,b,c under which the two roots are equal.
Understand the problem
`Nature of the roots' means describing what kind of roots the equation has --- real or not, rational or not --- without necessarily writing out their exact values. First we need to work out what can be said about the roots in general (given that a,b,c are rational numbers), then separately find the extra condition needed for the two roots to collapse into one repeated root.
The quadratic (a+2b−3c)x2+(b+2c−3a)x+(c+2a−3b)=0 has rational a,b,c.
The question asks what can be said about the nature of the roots, and then for the condition on a,b,c that makes the two roots equal.
Checking the discriminant directly leads to a large expansion. Notice something simpler first: what do the three coefficients add up to?
Add the three coefficients together. The cyclic pattern is designed to make things cancel -- see what the sum simplifies to, and what that tells you about x=1. Once you know one root, the sum-of-roots relation gives the other. Equal roots then becomes a condition relating the two.
Adding the three coefficients:
(a+2b−3c)+(b+2c−3a)+(c+2a−3b)=(a−3a+2a)+(2b+b−3b)+(−3c+2c+c)=0.
Since the coefficients of a quadratic sum to zero exactly when x=1 is a root, x=1 is always a root.
Since a,b,c are rational, all three coefficients are rational, so the sum of the roots, a+2b−3c3a−b−2c, is rational. With one root equal to 1, the other root, their sum minus 1, is also rational:
a+2b−3c3a−b−2c−1=a+2b−3c3a−b−2c−(a+2b−3c)=a+2b−3c2a−3b+c.
Hence both roots are rational: x=1 and x=a+2b−3c2a−3b+c.
For the two roots to be equal, the second root must also equal 1:
a+2b−3c2a−3b+c=1⟹2a−3b+c=a+2b−3c⟹a−5b+4c=0⟹a+4c=5b.
Let n be a positive integer. Find every integer a for which exactly n integers lie strictly between the two roots of x2+ax−1=0.
Understand the problem
`Strictly between the roots' means an integer bigger than the smaller root and smaller than the bigger one. Each value of a traps some number of integers this way. We need to find every integer a for which that count is exactly n.
x2+ax−1=0 has integer parameter a. Its two roots are always real.
Let n count the integers strictly between the roots. The question asks for every integer a that makes this count exactly n, for every positive integer n, as one formula in n.
Testing values of a one at a time won't work here. n can be any positive integer. There's no finite list to check.
Picture the two roots on a number line. An integer is trapped between them exactly when the quadratic is negative there.
Start at 0. Is 0 always trapped, no matter what a is? Check.
Now move outward, one integer at a time: 1,−1,2,−2,…. At some point on each side, the quadratic stops being negative. That point is the edge of the trap.
Find where that edge sits, in terms of a, for each distance from 0. Watch how the edge moves as the distance grows. That tells you exactly which integers are trapped for a given a -- and from there, which a traps exactly n of them.
Let f(x)=x2+ax−1.
An integer k is trapped strictly between the roots exactly when f(k)<0.
Evaluate f at x=0.
f(0)=
This is negative for every a, so 0 is always trapped.
For a positive integer m, evaluate f(m) and f(−m), and solve each for a.
f(m)<0⇒a<,f(−m)<0⇒a>
For a>0, check whether the first threshold can ever be cleared.
Positive integers are never trapped when a>0; only 0 and some negative integers can be.
The threshold m−1/m increases with m.
Let M be the largest m with m−1/m<a.
Total trapped count = (in terms of M)
Set this count equal to the target n and solve for M.
M=
Translate M=n−1 into bounds on a.
The largest m=n−1 must clear the threshold, but m=n must not.
<a≤
This interval has length 2, so exactly one integer can lie in it.
a=
Let f(x)=x2+ax−1. Since f opens upward, an integer k lies strictly between the two roots if and only if f(k)<0. The discriminant is a2+4>0 always, so the roots are real for every a.
At x=0: f(0)=−1<0 for every value of a, so 0 is always trapped.
For a positive integer m, evaluate f at ±m:
f(m)=m2+am−1<0f(−m)=m2−am−1<0⇒a<m1−m,⇒a>m−m1.
Suppose a>0. Since 1/m−m≤0 for every m≥1, the condition a<1/m−m can never hold. No positive integer is ever trapped when a>0.
So the trapped integers are exactly 0 together with whichever negative integers −m satisfy a>m−1/m.
The quantity m−1/m is strictly increasing in m. So the set of m satisfying m−1/m<a is exactly {1,2,…,M} for the largest such M.
Since 1−1/1=0<a always holds, M≥1.
The total trapped count is M+1.
Setting the count equal to a target positive integer n (so M=n−1, requiring n≥2) means m=n−1 must satisfy m−1/m<a, while m=n must not. That pins a to the interval
(n−1−n−11,n−n1].
This interval sits inside (n−2,n), and contains only the single integer n−1. So a=n−1 is the unique positive-integer solution.
Applying a→−a,x→−x turns this into the same problem, so a=−(n−1) is the unique negative-integer solution.
At n=1: neither branch applies, so check directly. a=0 traps only 0 (count 1).
Any a=0 traps at least one of ±1 in addition to 0 (count ≥2).
So a=0 is the unique solution, consistent with the general formula since ±(n−1)=0 at n=1.
Suppose α,β are the roots of ax2+bx+c=0, and that γ and −β are the roots of Ax2+Bx+C=0. Find the equation whose roots are α and γ.
Understand the problem
Two separate quadratics are given: one with roots α,β, another with roots γ and −β. We need to find the equation whose roots are α and γ --- one root taken from each original equation.
ax2+bx+c=0 has roots α,β. Ax2+Bx+C=0 has roots γ,−β.
The question asks for the equation with roots α,γ.
Solving for α,β,γ individually is underdetermined. Only the sum and product of α,γ are needed.
The sum falls out directly: β cancels once the sum-of-roots relations from both equations are combined. The product is harder: a direct attempt leaves a leftover α. Apply the same sum-of-roots idea to the reciprocals of the roots instead, since reciprocal roots correspond to reversing a quadratic's coefficients. That gives the sum of 1/α+1/γ, which relates to the product via the already-known sum.
Let f(x)=x2+2bx+1. The equation 2f(x)+ν(x2+1)=0 has real roots whenever f(x) has complex roots, and complex roots whenever f(x) has real roots. Find ν, in terms of b.
Understand the problem
We combine f with an extra term scaled by an unknown ν to build a new equation. Whenever f's roots are complex the new one's are real, and vice versa. We need to find the one value ν, in terms of b, that makes this swap always happen.
f(x)=x2+2bx+1 is given. Build a second equation 2f(x)+ν(x2+1)=0 for some multiple ν.
The question asks for ν, in terms of b, so the second equation's roots always swap character with f's: real whenever f is complex, and complex whenever f is real.
Expand the second equation directly, and write its discriminant as a function of b and ν.
Factor both discriminants into linear factors in b.
For one to always have opposite sign to the other, arrange for the second discriminant to equal −1 times a perfect square times the first, since a square is never negative. Match the linear factors on each side through a common multiplier, and solve for that multiplier -- that pins down ν.
f(x)=x2+2bx+1, discriminant D1=4(b−1)(b+1). Seek ν, in terms of b, so that 2f(x)+ν(x2+1)=0 always swaps real↔complex with f
Expand the second equation, compute its discriminant.
D2=
Factor D2 as a difference of two squares, in (2+ν) and 2b.
D2=−4()()
For opposite sign to be guaranteed for every b, force D2=−k2D1: match each factor of D2 to a factor of D1, through the same k.
2+ν−2b=k(), \quad 2+ν+2b=k()
Subtract the two equations to solve for k.
k=
Substitute back to solve for ν, then verify D2=−k2D1 directly.
ν=
Write f(x)=x2+2bx+1, with discriminant
D1=4(b2−1)=4(b−1)(b+1).
The second equation is
2f(x)+ν(x2+1)=(2+ν)x2+4bx+(2+ν)=0,
with discriminant
D2=16b2−4(2+ν)2=−4[(2+ν)2−(2b)2]=−4(2+ν−2b)(2+ν+2b).
We need D1 and D2 to have opposite sign for every b: real roots of one exactly when the other has complex roots. A square is never negative, so if D2=−k2D1 for some expression k, opposite sign is guaranteed automatically for every b at once -- no need to check sign case by case.
To force D2=−k2D1, match D2's two linear factors to D1's two linear factors (b−1) and (b+1), through the same multiplier k:
2+ν−2b=k(b−1),2+ν+2b=k(b+1).
Then, as required:
D2=−4⋅k(b−1)⋅k(b+1)=−k2⋅4(b−1)(b+1)=−k2D1.
Subtracting the two equations:
4b=k(b+1)−k(b−1)=2k,
so k=2b.
Substituting back into the first equation:
2+ν−2b=2b(b−1)=2b2−2b,
so ν=2b2−2.
Check directly: with ν=2b2−2, so 2+ν=2b2,
D2=16b2−4(2b2)2=16b2−16b4=16b2(1−b2).
Since b2≥0, this has sign exactly opposite to (b2−1) -- hence opposite to D1=4(b2−1) -- for every b=0. So real roots of f force complex roots of the second equation, and complex roots of f force real roots of the second equation.
Let f(x)=ax2+2bx+c, g(x)=a′x2+2b′x+c′ and h(x)=(b2−ac)x2+(2bb′−ac′−a′c)x+(b′2−a′c′). If h(x)=0 has equal roots, prove that f(x)=0 and g(x)=0 must share a common root.
Understand the problem
h is a quadratic built from the coefficients of two other quadratics f,g. We're told h has a repeated root, and we need to prove that this forces f,g to share a root too.
at2+2bt+c=0 and a′t2+2b′t+c′=0 are given. h(x)=(b2−ac)x2+(2bb′−ac′−a′c)x+(b′2−a′c′) has a repeated root.
The question asks to show the two original equations must share a root.
No candidate root is given, only that h's discriminant vanishes. An actual value has to be produced and checked.
Treat the two original equations not as quadratics in t, but as linear equations in the two unknowns t2 and t. Solve that linear system by elimination. This gives candidate values for t2 and t separately. Check they are consistent, i.e.\ that the candidate for t, squared, equals the candidate for t2: this holds exactly when h's discriminant vanishes.
h(x)=(b2−ac)x2+(2bb′−ac′−a′c)x+(b′2−a′c′) has equal roots
Expand the discriminant of h, collect, and set it to zero.
(ac′−a′c)2=4(ab′−a′b)()
Treat at2+2bt+c=0,a′t2+2b′t+c′=0 as linear equations in the unknowns t2 and t. Solve by elimination.
Candidate t=, candidate t2= (in terms of the six coefficients)
Check that (candidate t)2= candidate t2, using the identity above.
Hence, this t satisfies both at2+2bt+c=0 and a′t2+2b′t+c′=0: a genuine common root
Let h(x)=(b2−ac)x2+(2bb′−ac′−a′c)x+(b′2−a′c′). Its discriminant is
(2bb′−ac′−a′c)2−4(b2−ac)(b′2−a′c′).
Expanding both terms and collecting shows this equals
(ac′−a′c)2−4(ab′−a′b)(bc′−b′c).
Since h has equal roots, this discriminant is zero:
(ac′−a′c)2=4(ab′−a′b)(bc′−b′c).(25.1)
Treat at2+2bt+c=0 and a′t2+2b′t+c′=0 not as quadratics in t, but as a pair of linear equations in the two unknowns T:=t2 and U:=t:
aT+2bU=−c,a′T+2b′U=−c′.
Assuming ab′−a′b=0, this linear system has the unique solution:
U=2(ab′−a′b)a′c−ac′,T=ab′−a′bbc′−b′c.
By construction, this (T,U) satisfies both linear equations.
One thing is still unconfirmed: does U2=T? Check, using (25.1):
U2=(2(ab′−a′b)a′c−ac′)2=4(ab′−a′b)2(ac′−a′c)2.
Using (25.1), the numerator (ac′−a′c)2=4(ab′−a′b)(bc′−b′c), so
U2=4(ab′−a′b)24(ab′−a′b)(bc′−b′c)=ab′−a′bbc′−b′c=T.
So t:=U satisfies t2=T. Substituting T=t2,U=t into the linear system directly gives at2+2bt+c=0 and a′t2+2b′t+c′=0 -- t satisfies both original quadratics. Hence t=2(ab′−a′b)a′c−ac′ is an explicit common root of f(x)=0 and g(x)=0.
Let A,B,C be the angles of a triangle, and suppose tanBtanC=p. Solve for the range of p in terms of A.
Understand the problem
A,B,C are a triangle's angles (so they add to 180°). A is fixed. For example, if A=60°, then B+C=120°, and B,C can range over any split of that 120° that keeps both angles positive. As B,C vary, tanBtanC changes. We need to describe every value this product p can take, in terms of A.
A triangle has angles A,B,C, with A fixed. Let p=tanBtanC.
The question asks for the range of p, in terms of A.
Only B+C=π−A is fixed, leaving one degree of freedom in how the sum splits. As that split varies, p traces a curve in one variable.
Use the product-to-sum identities to rewrite p as a ratio depending on B,C only through cos(B−C) --- collapsing the one remaining degree of freedom into a single bounded quantity. As the split varies, that quantity sweeps a known interval, reaching one end exactly at the symmetric split B=C. Check whether p increases or decreases as this quantity moves through that interval, then read off the range from its value at the reachable endpoint. Whether the range is bounded or has a gap depends on whether A is acute, right, or obtuse.
Write θ=π−A; since A+B+C=π, this is also B+C, fixed once A is fixed. Using the product-to-sum identities 2sinBsinC=cos(B−C)−cos(B+C) and 2cosBcosC=cos(B−C)+cos(B+C), and writing u=cos(B−C) and
k=cosθ=cos(π−A)=−cosA,
we get:
p=tanBtanC=cosBcosCsinBsinC=u+ku−k.
As B ranges over (0,θ) with C=θ−B, the quantity B−C=2B−θ ranges over (−θ,θ), so u=cos(B−C) ranges over (k,1], as Figure~\figref{D26urange} shows. The value 1 is attained exactly at B=C=θ/2.
The endpoint k itself is excluded, since B,C>0 strictly.
Rewrite p=u+ku−k as p=1−u+k2k, by adding and subtracting k in the numerator.
Let w:=u+k. For 0<w1<w2, cross-multiplying gives 1/w1>1/w2; the same argument holds for w1<w2<0. So 1/w strictly decreases as w increases, on each side of w=0 separately.
Hence p=1−2k/w moves opposite to 1/w: p increases with u when k>0, and decreases with u when k<0.
The sign of k=−cosA splits the problem into three cases: A>π/2 (so k>0), A=π/2 (so k=0), and A<π/2 (so k<0).
Case 1 (A>π/2, k>0, p increasing in u). As u→k+, p→0.
At u=1:
p=1+k1−k=1−cosA1+cosA=cot2(A/2),
and this value is attained.
So p∈(0,cot2(A/2)].
Case 2 (A=π/2, k=0). Then p=u/u, which is 1 identically.
Case 3 (A<π/2, k<0, p decreasing in u). There is a pole at u=−k, i.e.\ at u=cosA, which lies in (k,1).
On u∈(k,−k): p decreases from 0− down to −∞, giving p∈(−∞,0).
On u∈(−k,1]: p decreases from +∞ down to cot2(A/2) at u=1, giving p∈[cot2(A/2),∞).
In an exam, when solving x2+px+q=0, a student miscopies the coefficient of x as p1 and obtains roots α1,β1. A second student miscopies the constant term as q2 and obtains roots α2,β2. Find the roots of the original equation, in terms of α1,β1,α2,β2.
Understand the problem
Two students each miscopy the same equation differently --- one gets the x-coefficient wrong, one gets the constant wrong --- and each solves their own wrong version, getting roots α1,β1 and α2,β2. We need to recover the true original roots from just these four numbers.
Two students each solved the same quadratic x2+px+q=0, each copying one coefficient correctly and one incorrectly. The two faulty solves gave two different pairs of roots.
The question asks for the true roots, using only the two faulty pairs.
Neither faulty solve alone recovers the true equation: each is consistent with only one correctly-copied coefficient.
For each faulty solve, identify which of the two root relations, sum or product, survived untouched, since it depends only on the coefficient copied correctly. Extract that one relation from each pair. This supplies both true coefficients independently, and the quadratic formula finishes the job.
Student 1 copies the constant term correctly but miscopies the coefficient of x as p1. Solving x2+p1x+q=0 gives roots α1,β1. Since only the linear coefficient was disturbed, the product of roots, which depends only on the constant term, still equals the true value:
q=α1β1.
Student 2 copies the coefficient of x correctly but miscopies the constant term as q2. Solving x2+px+q2=0 gives roots α2,β2. Since only the constant term was disturbed, the sum of roots, which depends only on the linear coefficient, still equals the true value:
p=−(α2+β2).
Rebuilding the true equation from these two independently-recovered coefficients:
x2−(α2+β2)x+α1β1=0.
Hence the true roots are
x=2(α2+β2)±(α2+β2)2−4α1β1.
Determine the condition for the roots of the following equation to be real: (a−b)2x2+2(a+b−2c)x+m2=0(a<b).
Understand the problem
We simply need to find the condition on a,b,c,m guaranteeing real roots here. We're already told a<b. The only thing left to determine is what else must hold.
The question asks for the condition on a,b,c,m under which this quadratic has real roots.
The discriminant, computed directly, is a difference of two squared expressions. A difference of squares always factors, without needing to expand either square.
Divide the discriminant by 4 and write it as that difference of squares. Factor it. Real roots require the product non-negative, which reduces to an absolute-value inequality.
Find the number of quadratic equations that are unchanged by raising their roots to the nth power, for a fixed integer n≥2, in terms of n.
Understand the problem
`Unchanged by raising roots to the nth power' means: raise both roots to that power, and get back exactly the same pair (possibly swapped). We need to count how many quadratics have this property, for a given n.
A quadratic's two roots r,s are raised to a fixed power n. Consider when {rn,sn}={r,s}.
The question asks how many such quadratic equations exist, as a function of n.
Restricting to real roots misses complex solutions. Count both. There are two structurally different ways for the pair to map to itself: each root fixed individually, or the two roots swapped.
In case (a), each root individually returns to itself under the nth power, giving n candidate values. In case (b), the two roots swap under the nth power, so applying the power twice must return a root to itself. Solve that broader condition, discard the case-(a) overlap, and pair up what remains.
Let the two roots be r,s. ``Unchanged by raising its roots to the nth power'' means {rn,sn}={r,s} as a set, which splits into two cases: each root maps to itself, or the two roots swap.
Case (a): each root maps to itself.rn=r and sn=s.
The equation zn=z, i.e.\ z(zn−1−1)=0, has solutions z=0 together with the n−1 complex (n−1)th roots of unity: n candidate values in total.
Taking r,s as an unordered pair from these n candidates, allowing repetition, the number of such pairs is
(2n+1)=2n(n+1).
Case (b): the roots swap.rn=s, sn=r, with r=s.
Substituting one relation into the other:
rn2=(rn)n=sn=r,
so r(rn2−1−1)=0.
If r=0, then s=rn is also 0, so r and s are both 0 -- contradicting r=s. So r=0, forcing rn2−1=1.
Since n−1 divides n2−1=(n−1)(n+1), every case-(a) candidate with zn−1=1 automatically satisfies zn2−1=1 too.
Discarding these n−1 overlapping candidates from the n2−1 roots of zn2−1=1 leaves exactly n2−1−(n−1)=n2−n genuine case-(b) candidates.
Each such r pairs with a distinct s=rn=r, also a case-(b) candidate. So these n2−n values split into 2n2−n unordered pairs, each giving one equation.
Let α,β be the roots of λ2(x2−x)+2λx+3=0. Suppose λ1,λ2 are the values of λ for which βα+αβ=k, where k>0. Find the equation whose roots are λ12/λ2 and λ22/λ1.
Understand the problem
This equation's coefficients depend on λ, so different λ give different roots α,β. The combination α/β+β/α hits a fixed target k at exactly two values λ1,λ2. We need to build the equation whose roots come from those two.
λ2(x2−x)+2λx+3=0 has roots α,β depending on λ. The condition βα+αβ=k picks out two values λ1,λ2.
The question asks for the equation with roots λ12/λ2 and λ22/λ1.
This is two layers: first find which λ satisfy the condition, then build the equation from those values. Don't shortcut past the first layer.
Express the condition in terms of the original equation's coefficients (which depend on λ), giving a quadratic in λ satisfied by λ1,λ2. From its sum and product, build the sum of cubes λ13+λ23, and hence the sum and product of the target roots, without solving for λ1,λ2 explicitly.
Writing λ2(x2−x)+2λx+3=0 as λ2x2+λ(2−λ)x+3=0, the coefficient-root relations give
α+β=λλ−2=1−λ2,αβ=λ23.
Then
α2+β2=(α+β)2−2αβ=1−λ4−λ22.
Imposing αβα2+β2=k:
λ231−λ4−λ22=k⟹λ2−4λ−2=3k⟹λ2−4λ−(2+3k)=0.
So λ1,λ2 are the roots of this quadratic: λ1+λ2=4, λ1λ2=−(2+3k).
The target roots are λ12/λ2 and λ22/λ1. Their product is λ1λ2=−(2+3k). Their sum is
λ1λ2λ13+λ23=λ1λ2(λ1+λ2)3−3λ1λ2(λ1+λ2)=−(2+3k)64+12(2+3k)=−2+3k88+36k.
Clearing denominators, the equation with these roots is
(2+3k)x2+(88+36k)x−(2+3k)2=0,
valid for real k>−2, k=−32, which keeps λ1,λ2 real, distinct, and nonzero.
Let m≥2 be an integer. Suppose a,b,c∈N and ax2−bx+c=0 has two distinct roots in (0,1), the larger being m times the smaller. Find the least possible value of a, in terms of m.
Understand the problem
a,b,c are positive integers, both roots in (0,1), the larger exactly m times the smaller. We need the smallest possible a.
ax2−bx+c=0 has natural-number coefficients and two distinct real roots α<β in (0,1), with β=mα for a given positive integer m.
The question asks for the smallest possible a, in terms of m.
``Both roots in (0,1)'' and ``ratio exactly m'' are not independent: the ratio condition is exact, and combined with the coefficient-root relations it collapses into one exact relation among a,b,c,m. Searching coefficients directly without that relation is unbounded.
Write α+β and αβ via the coefficient-root relations. Divide the product relation by the square of the sum relation to eliminate α, giving one equation in a,b,c,m. A coprimality argument then forces b to be a multiple of m+1, reducing the search to one integer parameter. Check its smallest values by hand.
Roots α<β=mα, both in (0,1). Write S:=α+β=α(1+m), P:=αβ=mα2.
Read off S,P from the coefficients.
S=,P=
Divide P by S2 to eliminate α.
mb2=
m and 1+m are consecutive integers, so their greatest common divisor -- the largest number dividing both, e.g.\ gcd(8,12)=4 -- is exactly 1, forcing 1+m to divide b.
b=(1+m)k, for a positive integer k
Substitute this into (31.1) and simplify.
ac=
Translate β<1 into a condition on a,k,m.
a≥≤mk
Check k=1: ac=m forces a to divide m, but the interval constraint needs a>m.
triple exists at k=1
Check k=2: ac=4m, a>2m.
Smallest such a forces c=, a=
Check k≥3 can't beat this: repeat the k=2 argument for general k.
>4m exactly when (k−)2>0, true for every k≥3
Least possible a is , attained uniquely at (a,b,c)=
Let the roots be α<β, with β=mα, both in (0,1), and write S:=α+β, P:=αβ. By the coefficient-root relations,
S=α(1+m)=ab,P=mα2=ac.
Dividing P by S2 eliminates α entirely:
S2P=α2(1+m)2mα2=(1+m)2m⟹(b/a)2c/a=(1+m)2m⟹mb2=ac(1+m)2.(31.1)
This is an exact equation: for a natural-number triple (a,b,c) to give roots in ratio exactly m, (31.1) must hold exactly.
Reducing (31.1).m and 1+m are consecutive integers, so their greatest common divisor -- the largest number dividing both, e.g.\ gcd(8,12)=4 -- is exactly 1. So (1+m)2 shares no factor with m, meaning (1+m)2 must divide b2 outright, which forces 1+m to divide b. Write b=(1+m)k for a positive integer k. Substituting into (31.1):
m(1+m)2k2⟹ac=ac(1+m)2=mk2.(31.2)
Why k=1 fails. At k=1: (31.2) becomes ac=m, so a must divide m. But the interval constraint needs a>m, and no positive divisor of m can exceed m itself. No valid triple exists at k=1.
Why k=2 gives the minimum, a=4m. At k=2: (31.2) becomes ac=4m, with a>2m required. Any divisor a of 4m satisfying a>2m must have c=4m/a<2, forcing c=1 and a=4m. Since 4m>2m holds for every m≥2, this is achievable, and it is the only solution at k=2.
Why k≥3 cannot beat this. At general k, (31.2) gives ac=mk2 with a>mk. Writing a=mk2/c for integer c≥1, the constraint a>mk becomes c<k. The smallest such a occurs at c=k−1, giving a=mk2/(k−1). Since k−1k2>4 exactly when (k−2)2>0, true for every k≥3, this exceeds 4m. So every k≥3 forces a strictly above 4m.
The least possible a is therefore 4m, attained uniquely at k=2, where b=(1+m)k=2m+2 and c=1. The roots are α=b/(a(1+m))=2m1 and β=mα=21.
x2+cx is given, with c>0 free, and x ranging over all reals.
The question asks for c so that the range of this fraction is exactly [−m,m].
Set the fraction equal to a target value y. Cross-multiply into a quadratic in x. Ask for which y that quadratic has a real solution. The discriminant condition gives the range, with both endpoints attained. Match that range's boundary against [−m,m].
Set y=x2+cx and cross-multiply, collecting as a quadratic in x:
yx2−x+yc=0.
For y=0, this forces x=0 directly, a genuine solution. For y=0, a real x exists exactly when the discriminant is non-negative:
1−4y2c≥0⇒y2≤4c1⇒−2c1≤y≤2c1.
So the range of x/(x2+c) is exactly [−2c1,2c1], both endpoints attained at x=±c.
For this range to equal [−m,m] exactly, set
2c1⟹c⟹c=m=2m1=4m21.
Let α,β,γ be the angles of a triangle. Three quadratic equations are given:
\begin{gather*}
x^2+\sin\alpha\,x+\cos\alpha=0,\\
x^2+\sin\beta\,x+\cos\beta=0,\\
x^2+\sin\gamma\,x+\cos\gamma=0,
\end{gather*}
and each pair of them shares a root. Prove that
\begin{align*}
&5+2\cos\alpha\cos\beta\cos\gamma+12\sin\frac\alpha2\sin\frac\beta2\sin\frac\gamma2\\
&\quad=\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha).
\end{align*}
Understand the problem
Three quadratics are given, from the triangle's angles. Every pair shares a root; we need to prove this forces the stated identity in α,β,γ.
Three quadratics are built from sin,cos of a triangle's three angles, and every pair of them shares a root.
The question asks you to prove one specific trigonometric identity in α,β,γ that this forces.
Name the three shared roots, and read off each equation's sine and cosine coefficient via the coefficient-root relations. Square and add the sine relations, add the cosine relations, and combine them. Separately, expand and add the pairwise products of the sines. Both routes land on the same expression in the shared roots, giving one identity relating sin2α+sin2β+sin2γ, cosα+cosβ+cosγ, and sinαsinβ+sinβsinγ+sinγsinα -- true for any three quadratics that pairwise share roots, before ever using that the angles form a triangle.
Now bring in α+β+γ=π. Each of those three symmetric sums has a classical closed form once the angles sum to π. The sine-squares sum reduces to a multiple of cosαcosβcosγ, the cosine sum reduces to a product of half-angle sines, and the sine-product sum reduces to a combination of the cosine sum and the cosines of the pairwise angle differences. Substituting all three into the identity from before, and simplifying, gives the result.
Three quadratics, built from sin,cos of a triangle's angles α,β,γ, pairwise share roots. Prove the identity this forces
Name the shared roots a,b,c (first-second, second-third, third-first). Read off each equation's coefficients via the coefficient-root relations.
sinα=,cosα= (and cyclically for β,γ)
Square and add the three sine relations.
sin2α+sin2β+sin2γ= (in terms of a,b,c)
Add the three cosine relations, then combine with the result above.
sin2α+sin2β+sin2γ+4(cosα+cosβ+cosγ)=
Separately, expand and add the three pairwise products of the sines.
sinαsinβ+sinβsinγ+sinγsinα= (in terms of a,b,c)
Doubling this matches the combined result above, giving one identity true for any three such quadratics.
Using γ=π−α−β, simplify sin2α+sin2β+sin2γ via cos2α+cos2β+cos2γ+2cosαcosβcosγ=1.
sin2α+sin2β+sin2γ=
Simplify cosα+cosβ+cosγ by sum-to-product, using sin2γ=cos2α+β.
cosα+cosβ+cosγ=
Simplify sinαsinβ+sinβsinγ+sinγsinα by product-to-sum.
(in terms of cosα+cosβ+cosγ and the pairwise angle-difference cosines)
Substitute all three simplifications into the identity from before, and simplify.
Let a be the root shared by the first and second equations, b the root shared by the second and third, and c the root shared by the third and first. The first equation's roots are then c,a; the second's are a,b; the third's are b,c. By the coefficient-root relations,
sinαcosαsinβcosβsinγcosγ=−(a+c),=ac,=−(a+b),=ab,=−(b+c),=bc.
Square and add the three sine relations:
sin2α+sin2β+sin2γ=(a+c)2+(a+b)2+(b+c)2=2a2+2b2+2c2+2ab+2bc+2ca.
Add the three cosine relations:
cosα+cosβ+cosγ=ab+bc+ca.
Combine these two: adding 4(cosα+cosβ+cosγ) to sin2α+sin2β+sin2γ gives
2(a2+b2+c2)+6(ab+bc+ca).
Expand the products sinαsinβ,sinβsinγ,sinγsinα:
sinαsinβsinβsinγsinγsinα=(a+c)(a+b)=a2+ab+ac+bc,=(a+b)(b+c)=ab+ac+b2+bc,=(b+c)(a+c)=ab+ac+bc+c2.
Add these three:
sinαsinβ+sinβsinγ+sinγsinα=(a2+b2+c2)+3(ab+bc+ca).
Doubling gives 2(a2+b2+c2)+6(ab+bc+ca), matching the combined result from before.
So pairwise root-sharing forces exactly
sin2α+sin2β+sin2γ+4(cosα+cosβ+cosγ)=2(sinαsinβ+sinβsinγ+sinγsinα).(33.1)
Since α,β,γ are a triangle's angles, γ=π−α−β, so cosγ=−cos(α+β). Also sin2γ=cos(2α+β).
Simplifying sin2α+sin2β+sin2γ. Using cos2θ=21+cos2θ:
cos2α+cos2β=21+cos2α+21+cos2β=1+2cos2α+cos2β.
By sum-to-product, cos2α+cos2β=2cos(α+β)cos(α−β), so
cos2α+cos2β=1+cos(α+β)cos(α−β)=1−cosγcos(α−β).
Also, using 2cosαcosβ=cos(α−β)+cos(α+β):
2cosαcosβcosγ=cosγcos(α−β)+cosγcos(α+β)=cosγcos(α−β)−cos2γ.
Adding cos2γ to the last two results cancels the cosγcos(α−β) and −cos2γ terms:
cos2α+cos2β+cos2γ+2cosαcosβcosγ=1.
So sin2α+sin2β+sin2γ=3−(cos2α+cos2β+cos2γ) becomes
sin2α+sin2β+sin2γ=2+2cosαcosβcosγ.
Simplifying cosα+cosβ+cosγ. By sum-to-product, using 2α+β=2π−2γ:
cosα+cosβ=2cos(2α+β)cos(2α−β)=2sin2γcos(2α−β).
Since sin2γ=cos(2α+β), the bracket is a difference of cosines:
cos(2α−β)−cos(2α+β)=2sin2αsin2β.
So
cosα+cosβ+cosγ=1+4sin2αsin2βsin2γ.
Simplifying sinαsinβ+sinβsinγ+sinγsinα. By product-to-sum,
sinαsinβ=21[cos(α−β)−cos(α+β)]=21[cos(α−β)+cosγ],
and cyclically for the other two products.
Adding all three:
sinαsinβ+sinβsinγ+sinγsinα=21(cosα+cosβ+cosγ)+21cos(α−β)+21cos(β−γ)+21cos(γ−α).
Substitute the three simplifications into (33.1). The left side becomes
2+2cosαcosβcosγ+4[1+4sin2αsin2βsin2γ]=6+2cosαcosβcosγ+16sin2αsin2βsin2γ.
The right side becomes
2[21+2sin2αsin2βsin2γ+21(cos(α−β)+cos(β−γ)+cos(γ−α))]=1+4sin2αsin2βsin2γ+cos(α−β)+cos(β−γ)+cos(γ−α).
Setting the two sides equal:
5+2cosαcosβcosγ+12sin2αsin2βsin2γ=cos(α−β)+cos(β−γ)+cos(γ−α).
Suppose f(x) is defined by: (f2cosθ+fcos2θ+cos3θ)x+(f2sinθ+fsin2θ+sin3θ)=0. If f(x) is a rational function of x for every x, then investigate the permissible values of θ.
Understand the problem
This is a quadratic in f, with x,θ building its coefficients. ``Rational'' means f is a ratio of polynomials in x, though those coefficients can be irrational. We need which θ make this true.
(f2cosθ+fcos2θ+cos3θ)x+(f2sinθ+fsin2θ+sin3θ)=0 is given.
For every x, this can be solved for f using only rational operations, no square roots, and f must come out real, not complex. The question asks what values θ can take.
Solving for f means applying the quadratic formula in f, treating x as a parameter. Avoiding square roots is a statement about that quadratic's discriminant, viewed as a polynomial in x -- but a discriminant that's a perfect square can still be a negative one, which would make f complex instead of real. Both possibilities need ruling out separately.
Treat the relation as a quadratic in f with coefficients linear in x. Write its discriminant, quadratic in x. Two things must hold: that discriminant must be a perfect square in x, and its leading coefficient must be non-negative. Demanding the perfect-square condition, after substituting the multiple-angle values and simplifying each cross-term via the sine-subtraction formula, reduces to one trigonometric equation with two candidate families of θ. Check the leading-coefficient condition on each family separately to see which one actually survives.
View (f2cosθ+fcos2θ+cos3θ)x+(f2sinθ+fsin2θ+sin3θ)=0 as a quadratic in f.
(xcosθ+sinθ)f2+(xcos2θ+sin2θ)f+(xcos3θ+sin3θ)=0
Write its discriminant as a polynomial in x.
D(x)=x2+x+
For f to be real and rational in x, two conditions on D(x) must both hold.
Condition 1: D(x) is a perfect square in x. Condition 2: its leading coefficient is
Using shorthand a=cosθ,b=cos2θ,c=cos3θ,a′=sinθ,b′=sin2θ,c′=sin3θ, derive Condition 1 by demanding D(x)'s own discriminant vanish.
That condition reduces to ()2=()()
Simplify ab′−a′b and bc′−b′c via the sine-subtraction formula.
ab′−a′b=,bc′−b′c=
Simplify ac′−a′c the same way.
ac′−a′c=
Solve Condition 1, now written purely in sinθ,cosθ.
sinθ=0 or cosθ=
Check Condition 2 on the sinθ=0 family: compute the leading coefficient b2−4ac there.
Leading coefficient =
Check Condition 2 on the other family the same way.
Leading coefficient =
Only the family with a non-negative leading coefficient survives Condition 2, giving θ=
View (f2cosθ+fcos2θ+cos3θ)x+(f2sinθ+fsin2θ+sin3θ)=0 as a quadratic in f, with x as a parameter:
(xcosθ+sinθ)f2+(xcos2θ+sin2θ)f+(xcos3θ+sin3θ)=0.
For brevity, write a=cosθ,b=cos2θ,c=cos3θ,a′=sinθ,b′=sin2θ,c′=sin3θ, so the equation reads (ax+a′)f2+(bx+b′)f+(cx+c′)=0.
Solving for f by the quadratic formula brings in D(x), where D(x) is this quadratic's discriminant, itself a polynomial in x:
D(x)=(bx+b′)2−4(ax+a′)(cx+c′)=(b2−4ac)x2+[2bb′−4(ac′+a′c)]x+(b′2−4a′c′).
For f to be a genuine real rational function of x -- not merely algebraic, and never complex -- two separate conditions must both hold.
Condition 1: D(x) must be a perfect square as a polynomial in x, so that D(x) is itself a polynomial in x. This means D(x)'s own discriminant, as a quadratic in x, must vanish.
Condition 2: a perfect square can still be a negative one -- D(x)=λ(x−x0)2 with λ<0 is negative for every x=x0, making D(x) imaginary almost everywhere. So the leading coefficient λ=b2−4ac must also be non-negative.
Deriving Condition 1: setting D(x)'s discriminant to zero,
[2bb′−4ac′−4a′c]2=4(b2−4ac)(b′2−4a′c′).
Dividing by 4 and expanding fully, this reduces to
(ac′−a′c)2=(ab′−a′b)(bc′−b′c).
Substituting back a=cosθ,b=cos2θ,c=cos3θ,a′=sinθ,b′=sin2θ,c′=sin3θ, each cross-term collapses via the sine-subtraction formula:
ab′−a′bbc′−b′cac′−a′c=cosθsin2θ−sinθcos2θ=sinθ,=cos2θsin3θ−sin2θcos3θ=sinθ,=cosθsin3θ−sinθcos3θ=sin2θ.
So Condition 1 becomes sin22θ=sin2θ.
Using sin2θ=2sinθcosθ:
4sin2θcos2θ=sin2θ⟹sin2θ(4cos2θ−1)=0⟹sinθ=0 or cosθ=±21.
Condition 1 alone allows two families: sinθ=0, or cosθ=±21. Condition 2 must now be checked on each.
Checking Condition 2 on sinθ=0, using cos2θ=2cos2θ−1 and cos3θ=4cos3θ−3cosθ:
If cosθ=1: then b=1 and c=1, so
λ=1−4(1)(1)=−3.
If cosθ=−1: then b=1 and c=−1, so
λ=1−4(−1)(−1)=−3 too.
Either way λ=−3<0: Condition 2 fails. The entire sinθ=0 family is spurious -- f is not real there.
Checking Condition 2 on cosθ=±21:
If cosθ=21: then b=−21 and c=−1, so
λ=41−4(21)(−1)=41+2=49.
If cosθ=−21: then b=−21 and c=1, so
λ=41−4(−21)(1)=41+2=49 too.
Either way λ=49>0: Condition 2 holds. This family survives.
Intersecting both conditions leaves only cosθ=±21, i.e.
θ=nπ±3π,n∈Z.
Let α,β be the roots of x2+px+q=0, and let m be a non-negative integer. Suppose α2m+2 and β2m+2 are the roots of x2−Rx+S=0. Show that x2−4q2mx+(2q2m+1−R)=0 always has two real roots, and find these roots in terms of p, q, and m.
Understand the problem
Squaring α,β repeatedly reaches the power 2m+2, the roots of a second equation. We need to show a third equation, from the first two, always has real roots, and solve it.
α,β are the roots of x2+px+q=0. After m+2 repeated squarings, α2m+2,β2m+2 are the roots of x2−Rx+S=0.
The question asks you to show that x2−4q2mx+(2q2m+1−R)=0 always has two real roots, and to write them explicitly in terms of p,q,m.
There is no way to reach α2m+2 from α in one step: it takes m+2 successive squarings, so a chain of intermediate power-sums has to be built one level at a time.
Define a running sum that adds the matching powers of α and β, with the exponent doubling at each step. Find a recursion connecting consecutive terms of this sequence: square a term and subtract off twice the corresponding power of q. Carry the recursion up to the level the problem asks about. This makes the target equation's discriminant collapse to a perfect square.
Define, for k≥0,
Tk:=α2k+β2k,
so T0=α+β, which is −p.
Squaring repeatedly, using α2k+1+β2k+1=(α2k+β2k)2−2(αβ)2k:
Tk+1=Tk2−2q2k,k=0,1,2,…
Each Tk is built purely from p,q this way.
Since α2m+2,β2m+2 are given as the roots of x2−Rx+S=0, their sum is R, i.e.\ R=Tm+2. By the recurrence:
Tm+2=Tm+12−2q2m+1.
Now examine x2−4q2mx+(2q2m+1−R)=0. Its discriminant is
16q2m+1−4(2q2m+1−R)=8q2m+1+4R=8q2m+1+4(Tm+12−2q2m+1)=4Tm+12,
four times a perfect square, always non-negative. So this equation always has two real roots, equal exactly when Tm+1=0.
By the quadratic formula:
x=24q2m±4Tm+12=2q2m±Tm+1.
So the two roots are 2q2m+Tm+1 and 2q2m−Tm+1, both computable from p,q,m alone via the recursion T0=−p,Tk+1=Tk2−2q2k.
At m=0: T1=p2−2q, and the two roots are 2q+(p2−2q)=p2 and 2q−(p2−2q)=4q−p2.
Deduce the condition for two quadratics to have a common root.
Understand the problem
`Sharing a common root' means one number solves both equations. We need the general rule, from the coefficients alone, plus the value of that shared root.
x2+px+q=0 and x2+rx+s=0 are two quadratics, assumed not identical.
The question asks for the condition on p,q,r,s under which the two equations share a root, and the value of that root.
Both equations are already monic, so a shared root can be isolated directly: subtracting one equation from the other cancels the leading term immediately, without needing either equation's own roots at all.
Subtract the two equations to get a linear equation in the shared root. Solve it, checking first that the coefficient you're dividing by can't vanish without making the two equations identical. Substitute this root back into either original equation, clear the resulting denominator, and simplify to get the condition on p,q,r,s alone.
Let the two equations be x2+px+q=0 and x2+rx+s=0, assumed not identical, and suppose they share a root x0.
Both are monic, so subtracting the second equation from the first cancels the x2 term outright:
(p−r)x0+(q−s)=0.
If p=r, this forces q=s too -- making the two equations identical, contradicting the assumption that they're not. So p=r whenever a genuine shared root exists, and
x0=p−rs−q.
Substitute x0 into the first equation, x02+px0+q=0, and clear the denominator (p−r)2:
(s−q)2+p(p−r)(s−q)+q(p−r)2=0.
Expand each product separately:
p(p−r)(s−q)q(p−r)2=p2s−p2q−prs+prq,=p2q−2pqr+qr2.
Add these two and simplify:
p(p−r)(s−q)+q(p−r)2=p2s−prs−pqr+qr2=ps(p−r)−qr(p−r)=(p−r)(ps−qr).
So the condition becomes (q−s)2+(p−r)(ps−qr)=0, i.e.\
(q−s)2=(p−r)(qr−ps).
Let m≥2 be an integer, and let fn(x)=(−1)n−1x2+pnx+q for n=1,…,m, where p1,…,pm,q are arbitrary real numbers. Investigate, in terms of m, the number of real roots that f1(x)f2(x)⋯fm(x)=0 is guaranteed to have.
Understand the problem
We build m quadratics sharing the same q, each with its own pn, with the squared term's sign flipping as n increases (e.g.\ f1=−x2+p1x+q, f2=x2+p2x+q). Multiplying them all together gives one big equation. We need to find how many real roots it's always guaranteed to have.
m quadratic-shaped factors, sharing the same constant term q but with the sign of the squared term alternating and middle coefficients pn free, are multiplied together.
The question asks for the guaranteed number of real roots of the whole product, as a formula in m.
A single factor's discriminant can always be made negative by its own free pn. No individual factor can be forced real-rooted. What can be forced is a guarantee across the whole alternating chain, by playing the sign of q against the alternating leading signs.
Rewrite every factor in monic form. This flips the discriminant's dependence on q differently for odd- and even-indexed factors. Whichever sign q has, it forces every discriminant of one parity to be non-negative.
Rewrite each factor in standard (monic) form. For odd n (leading sign +), fn=x2+pnx+q, discriminant Dn=pn2−4q. For even n (leading sign −), −x2+pnx+q=0 rewrites as x2−pnx−q=0, discriminant Dn=pn2+4q.
At least 2⌊m/2⌋ real roots, always.
If q≤0: every odd-indexed
Dn=pn2−4q=pn2+4∣q∣≥0,
so all ⌈m/2⌉ odd-indexed factors have two real roots each.
If q≥0: symmetrically, every even-indexed
Dn=pn2+4q≥0,
so all ⌊m/2⌋ even-indexed factors have two real roots each.
Every real q is ≥0 or ≤0, so at least min(⌈m/2⌉,⌊m/2⌋)=⌊m/2⌋ factors are guaranteed real-rooted, contributing at least 2⌊m/2⌋ real roots.
Let α,β be the roots of ax2+bx+c=0, and let f(x)=a1x2+b1x+c1. Find the equation whose roots are f(α) and f(β).
Understand the problem
α,β are one quadratic's roots, and f a separate quadratic expression. Plugging α,β into f gives two new numbers; we need the equation with these as roots.
ax2+bx+c=0 has roots α,β. f(x)=a1x2+b1x+c1 is a second, independent expression.
The question asks for the equation with roots f(α),f(β).
Substituting the roots' closed forms (each involving a square root) directly produces a mess. Instead, only f(α)+f(β) and f(α)f(β) are needed.
Use aα2+bα+c=0 to rewrite α2 as linear in α, and likewise for β. This makes f(α),f(β) linear in α,β respectively. Add and multiply these two linear expressions. Both reduce to combinations of α+β and αβ, already known from a,b,c.
Since α,β are roots of ax2+bx+c=0:
aα2+bα+c⟹α2=0=−abα+c,
and likewise for β.
Substitute into f(α)=a1α2+b1α+c1:
f(α)=a1(−abα+c)+b1α+c1=α(b1−aa1b)+(c1−aa1c)=aα(ab1−a1b)+(ac1−a1c).
Write M=ab1−a1b, N=ac1−a1c, so f(α)=aαM+N and likewise f(β)=aβM+N.
Adding, and using α+β=−b/a:
f(α)+f(β)=a(α+β)M+2N=a−abM+2N=a22aN−bM=a22a(ac1−a1c)−b(ab1−a1b).
Let pm be the m-th prime number (p1=2,p2=3,p3=5,…). One of the roots of a rational polynomial of degree 2m is
2−3+5−⋯+(−1)m+1pm.
Find the value of the smallest real root.
Understand the problem
p1,p2,… are the primes in order. We build one number by adding and subtracting the square roots of the first m primes, flipping sign each time. We need the smallest real root of the polynomial this number satisfies.
r=2−3+5−⋯+(−1)m+1pm is a root of a rational polynomial of degree 2m.
The question asks for the smallest root of that polynomial.
With m independent primes involved, no amount of repeated squaring isolates and clears the square roots in one or two steps. Instead, build up a rational polynomial having r as a root one prime at a time: pairing a shift by +p with a shift by −p and multiplying clears that square root out entirely, since a rational equation can never distinguish +p from −p.
That same sign-flip idea, applied directly to the given polynomial, shows it must contain every sign-flipped variant of r as a root too. Doing this for every prime gives 2m distinct numbers, exactly the polynomial's degree, so this is the entire root set. Pick out the smallest by pushing every term as low as possible at once.
r=∑i=1m(−1)i+1pi is a root of a rational polynomial of degree 2m. Build one prime at a time, starting from Q1(x):=x2−p1, with roots ±p1
Given Qk -- rational, with roots the 2k sign-combinations of the first k primes -- pair a shift by +pk+1 with a shift by −pk+1.
Qk+1(x):=Qk(x−pk+1)⋅Qk(x+pk+1)
Expand Qk(x−pk+1), separating even and odd powers of pk+1.
Qk(x−pk+1)=A(x)−B(x)pk+1, for rational A,B
Multiply this by Qk(x+pk+1)=A(x)+B(x)pk+1.
Qk+1(x)=, manifestly rational
Repeat this m−1 more times, to build Qm.
Qm has degree , with roots the 2m sign-combinations ∑iϵipi
A rational equation can't distinguish +pi from −pi, by the same even/odd-power split as above.
Flipping any one sign in a root of the given polynomial produces
Apply this one prime at a time to the given polynomial.
Its full root set is exactly , matching Qm's
Minimize each term ϵipi independently.
Smallest root needs ϵi= for every i at once
Let r=2−3+5−⋯+(−1)m+1pm be the given root, i.e.\ r=i=1∑m(−1)i+1pi.
Building a rational polynomial with r as a root, one prime at a time. Start with Q1(x):=x2−p1, which has roots ±p1.
Given Qk(x) -- rational, of degree 2k, with roots exactly the sign-combinations ∑i=1kϵipi, ϵi=±1 -- build the next one by pairing a shift by +pk+1 with a shift by −pk+1:
Qk+1(x):=Qk(x−pk+1)⋅Qk(x+pk+1).
Expand Qk(x−pk+1): every power of pk+1 that appears is either even -- giving a rational number, since (pk+1)2=pk+1 -- or odd -- giving a rational multiple of pk+1. Collecting these,
Qk(x−pk+1)=A(x)−B(x)pk+1,
for some polynomials A,B with rational coefficients. Flipping the sign of pk+1 only flips the sign of the odd-power part, so likewise
Qk(x+pk+1)=A(x)+B(x)pk+1.
Multiplying these two:
Qk+1(x)=A(x)2−pk+1B(x)2,
manifestly rational, since A,B,pk+1 all are.
Qk+1's roots are exactly x=±pk+1+(a root of Qk), since a root of either factor makes that factor vanish. This extends each of Qk's 2k sign-combinations by one more sign choice for pk+1, giving all 2k+1 sign-combinations of the first k+1 primes -- exactly matching Qk+1's degree, 2k+1.
Repeating this m−1 more times builds Qm(x): rational, of degree 2m, with r itself among its roots, and its full root set exactly the 2m sign-combinations ∑i=1mϵipi.
The given polynomial has this same root set. A rational-coefficient equation can't tell +pi apart from −pi: exactly as above, only even powers of pi ever produce a rational contribution, so substituting pi→−pi throughout turns any true rational-coefficient equation into another true one. So if a rational polynomial has r as a root, flipping any one sign in r must produce another root of that same polynomial.
Applying this one sign at a time, the given rational polynomial must have all 2m sign-combinations of r as roots. It has degree exactly 2m, and these 2m values are already distinct, so that accounts for every root: its full root set is precisely these 2m sign-combinations, matching Qm's.
Each term ϵipi is minimized on its own by taking ϵi=−1, since every pi>0, and every sign pattern genuinely occurs among the roots. So the smallest root is reached by flipping every sign to negative at once:
smallest root=−(2+3+5+⋯+pm)=−i=1∑mpi.
x2−3λx+(2λ2−m)=0 has λ>0 real. Its roots are real, and their product equals m2.
The question asks for every value of λ, in terms of m.
Solving the product condition alone pins λ2 to an expression in m, but doesn't confirm that expression is positive, nor that the equation's roots are genuinely real once plugged back in. Both checks are needed, separately.
Solve the product condition for λ2 in terms of m. Demand it comes out positive. Separately demand the discriminant be non-negative. Combine both requirements on m, then take the square root to recover λ itself.
The admissible values of m split into two disjoint pieces, with a trap between them.
Product of roots =2λ2−m. Set equal to target m2
Solve for λ2.
λ2= (in terms of m)
λ>0 requires this to be strictly positive.
m(m+1)>0 exactly when m>0 or m≥≤−1 \hfill (EXISTENCE)
Separately, demand the original equation's discriminant be ≥0.
Discriminant =
Simplify using λ2=m(m+1)/2.
≥0 exactly when m≥0 or m≥≤−9 \hfill (REALITY)
Combine EXISTENCE and REALITY --- the range −9<m<−1 satisfies existence but fails reality
The product of the roots of x2−3λx+(2λ2−m)=0 is 2λ2−m. Setting this equal to the given target m2:
2λ2−m⟹λ2=m2=2m2+m=2m(m+1).
For a real, positive λ to exist, we need λ2>0:
m(m+1)>0 exactly when m>0 or m<−1.(40.1)
Real-rootedness is not automatic and must be checked separately. The discriminant is
9λ2−4(2λ2−m)=λ2+4m=2m(m+1)+4m=2m(m+9),
so real roots require
m(m+9)≥0 exactly when m≥0 or m≤−9.(40.2)
Combine (40.1) and (40.2). For m>0: both hold. For −1<m≤0: (40.1) already fails. For m<−1: (40.1) holds, but (40.2) needs m≤−9. The intermediate range −9<m<−1 has a candidate λ that exists but fails to make the original quadratic's roots real. For m≤−9: both hold.
Hence λ=2m(m+1), valid exactly when m>0 or m≤−9.
Prove that the following equation has two real, distinct roots:
(x−a)(x−c)+k(x−b)(x−d)=0
where a,b,c,d,k∈R,a<b<c<d,k=−1
Understand the problem
a,b,c,d are four fixed numbers in increasing order, and k=−1 is fixed. We build a new quadratic by combining two products this way. We need to prove it always has two distinct real solutions, for every valid a,b,c,d,k.
a<b<c<d are real, and k=−1 is real. Let Q(x)=(x−a)(x−c)+k(x−b)(x−d).
The question asks you to show that, for every permitted k, Q always has two real, distinct roots.
With k not pinned to one value, expanding Q's discriminant gives no visibly fixed sign. The argument must hold for every k at once.
Evaluate Q at a,b,c,d themselves, rather than the discriminant directly. Two of these four values carry a fixed sign regardless of k. The other two flip sign with k. Split into k>0, k=0, k<0 and track the four signs in each case. A single strict sign change between two of the four points is already enough. A quadratic with real coefficients can only change sign between two real, distinct roots.
Fix real numbers a<b<c<d and a real number k=−1, and set
Q(x)=(x−a)(x−c)+k(x−b)(x−d).
The coefficient of x2 in Q is 1+k=0, so Q is a genuine quadratic.
Evaluate Q at the four given numbers. In Q(a) and Q(c), the first product vanishes automatically:
Q(a)Q(c)=(a−a)(a−c)+k(a−b)(a−d)=k(a−b)(a−d),=k(c−b)(c−d).
In Q(b) and Q(d), it is the second product that vanishes, leaving expressions with no k in them at all:
Q(b)Q(d)=(b−a)(b−c)+k(b−b)(b−d)=(b−a)(b−c),=(d−a)(d−c).
Pin down the signs. Since a<b<c<d:
(a−b)<0,(a−d)<0(c−b)>0,(c−d)<0(b−a)>0,(b−c)<0(d−a)>0,(d−c)>0⟹(a−b)(a−d)>0;⟹(c−b)(c−d)<0;⟹Q(b)<0;⟹Q(d)>0.
So Q(b)<0 and Q(d)>0 always. Since (a−b)(a−d)>0, Q(a) has the same sign as k; since (c−b)(c−d)<0, Q(c) has the opposite sign to k.
Why one sign change is enough.Q's real coefficients mean its two roots are either both real, or a complex-conjugate pair -- never one of each. A repeated root, or a complex pair, never lets Q strictly change sign anywhere. So a single strict sign change between any two points already forces Q's two roots to be real and distinct.
Case k>0. Signs at a,b,c,d are +,−,−,+. Q changes sign between a and b -- already enough.
Case k=0.Q(x)=(x−a)(x−c) outright, with roots a,c: real and distinct.
Case k<0. Signs at a,b,c,d are −,−,+,+. Q changes sign between b and c -- again enough on its own.
So for every real k=−1, Q always has two real, distinct roots.
abcd+−−+Case k>0: sign change on (a,b)abcd−−++Case k<0: sign change only on (b,c)
Suppose x+y+z=s and xy+yz+zx=p for real numbers x,y,z. Find, in terms of s and p, the least and greatest values that x can take.
Understand the problem
We have three unknowns x,y,z linked by only two equations, so many triples can satisfy both. Focusing on x, we need its full range: smallest and largest possible values, in terms of s,p.
The question asks for the smallest and largest values of x, in terms of s,p.
The system is two equations in three unknowns, underdetermined. Fixing x turns the remaining conditions into a statement about y,z alone: their sum and product both become known. Two reals with a given sum and product exist exactly when a standard inequality between that sum and product holds.
Treat x as fixed. Express y+z and yz in terms of x,s,p. Demand y,z real via (y+z)2≥4yz. This gives a quadratic inequality in x alone. Solve it.
Fix x and treat y,z as the remaining two unknowns. From the two given conditions:
y+z=s−x,yz=p−x(s−x)=p−sx+x2.
For y,z to be real, their sum and product must satisfy (y+z)2≥4yz:
(s−x)2≥4(p−sx+x2)⟹s2−2sx+x2≥4p−4sx+4x2⟹−3x2+2sx+s2−4p≥0,
i.e.
3x2−2sx−(s2−4p)≤0.
Solving 3x2−2sx−(s2−4p)=0 via the quadratic formula:
x=62s±4s2+12(s2−4p)=62s±4s2−3p=3s±2s2−3p.
Here s2−3p=21[(x−y)2+(y−z)2+(z−x)2]≥0 automatically, since s,p arise from actual real x,y,z.
Since the coefficient of x2 in 3x2−2sx−(s2−4p) is positive, the inequality holds exactly between these two roots. Hence
3s−2s2−3p≤x≤3s+2s2−3p.
If a,b,c are rational and a+b+c=0, then find the roots of:
(b+c−a)x2+(c+a−b)x+(a+b−c)=0.
Understand the problem
a,b,c are rational numbers satisfying one extra condition: they add up to zero. This equation's coefficients are built from a,b,c in a repeating pattern --- each one is two of the letters added together, minus the third. We simply need to solve this equation for x.
(b+c−a)x2+(c+a−b)x+(a+b−c)=0 has rational a,b,c with a+b+c=0.
The question asks what can be said about the nature of the roots, and then for the roots themselves.
The three coefficients, in this tangled form, obscure what's going on. Use the one unused fact: a+b+c=0, substituted directly into each coefficient.
Rewrite each coefficient using a+b+c=0. Each becomes a simple multiple of one quantity. The equation collapses to a much simpler quadratic, whose own coefficients happen to sum to zero. A quadratic whose coefficients sum to zero always has x=1 as a root, no discriminant needed. The other root then falls out of the product-of-roots relation.
Let a,b,c be real numbers, and suppose each pair of the equations ax2+bx+c=0, bx2+cx+a=0, cx2+ax+b=0 shares a common root. Investigate the constraint this puts on a,b,c.
Understand the problem
Three equations are built from the same three numbers a,b,c, cycled into a different order each time (first a,b,c, then b,c,a, then c,a,b). We're told every pair among these three equations shares a common root. We need to find what this forces a,b,c to satisfy.
ax2+bx+c=0, bx2+cx+a=0, cx2+ax+b=0 are given, with a,b,c cyclically rotated. Every pair shares a root.
The question asks what condition this forces on a,b,c.
Eliminating the shared root from each pair the standard way works, but doing it three times looks like a lot of algebra until you notice the three results share enough structure to combine directly.
Eliminate the shared root from each pair by cross-multiplication --- this gives three identities, one per pair, symmetric in a,b,c. Subtracting them pairwise factors nicely, forcing either the trivial case where all three are equal, or one genuine condition. The cyclic rotation itself then gives a shortcut to confirm that condition directly: evaluate each quadratic at x=1, where it collapses to the sum of its coefficients --- unchanged by cyclically rotating a,b,c --- so all three equations vanish together at x=1 the instant that condition holds.
For two quadratics A1x2+B1x+C1=0 and A2x2+B2x+C2=0 sharing a root t, the standard elimination gives
B1C2−B2C1t2=C1A2−C2A1t=A1B2−A2B11,
which in turn gives
(C1A2−C2A1)2=(B1C2−B2C1)(A1B2−A2B1).
Applying this to each pair among ax2+bx+c=0, bx2+cx+a=0, cx2+ax+b=0, and writing X=bc−a2,Y=ca−b2,Z=ab−c2:
X2=YZ,Y2=ZX,Z2=XY.
Subtracting the first two:
X2−Y2⟹(X−Y)(X+Y+Z)=YZ−ZX=−Z(X−Y)=0,
and cyclically for the other two pairs. So either X=Y=Z, or X+Y+Z=0.
Case X=Y=Z:
bc−a2⟹(b−a)(a+b+c)=ca−b2=0,
and cyclically. Taken together, either a=b=c, or a+b+c=0.
Case X+Y+Z=0: this reads a2+b2+c2=ab+bc+ca, i.e.\ 21[(a−b)2+(b−c)2+(c−a)2]=0, which for real a,b,c forces a=b=c, already covered above.
So the full condition is a=b=c or a+b+c=0. The a=b=c branch is degenerate: all three equations become identical, a(x2+x+1)=0. The genuine condition is
a+b+c=0.
Evaluating ax2+bx+c at x=1 gives exactly a+b+c, and this is true of all three equations, since cyclically rotating a,b,c never changes their sum. So the instant a+b+c=0, all three equations vanish at x=1 simultaneously.
Each equation's other root also follows directly, using the sum-of-roots relation. For ax2+bx+c=0, the roots sum to −b/a; since one root is 1, the other is −b/a−1=−(a+b)/a. Since a+b+c=0, this is c/a. Cyclically, bx2+cx+a=0's other root is a/b, and cx2+ax+b=0's other root is b/c.
Let m be a real number, and let a be an integer. In terms of m, find every non-positive integer a for which (a2−12)x4+mx2−4=0 has no real root at all.
Understand the problem
`No real root at all' means every one of the equation's four solutions is non-real --- not a single real number among them. Here m is a fixed real number, and a is a whole number that's zero or negative. We need to find every such a that makes this happen, written in terms of m.
(a2−12)x4+mx2−4=0 is given, with integer a and real m.
The question asks, in terms of m, for every non-positive integer a so that the equation has no real root at all.
Only even powers of x appear, so this is secretly a quadratic in x2. Don't assume the sign of m is irrelevant. For one sign of m, which values of a work doesn't depend on ∣m∣ at all. For the other sign, it does, and the set of working a's can even become empty.
Substitute u=x2. Look at the sum and product of this quadratic's two roots in u, not the discriminant first: whichever forces both roots negative decides everything, with a discriminant check needed as a final layer for one branch only.
Let u=x2. The equation becomes
(a2−12)u2+mu−4=0,
a genuine quadratic in u since a2=12 for any integer a. A real x exists exactly when this quadratic has a real root u≥0.
Write the sum and product of the two u-roots directly from the coefficients:
product=a2−12−4,sum=a2−12−m.
Branch 1: a2>12. Then a2−12>0, so the product is negative. A negative product means the two u-roots have opposite signs, so one is always positive -- giving a real x. So no a with a2>12 can ever give ``no real root,'' regardless of m.
Branch 2: a2<12. Then the product =a2−12−4>0: the two roots, whenever real, share a sign. Which sign is decided by the sum =a2−12−m.
If m≤0: −m≥0 and a2−12<0, so the sum ≤0. Whenever the roots are real, both are negative, so neither gives a real x. Every non-positive integer a with a2<12 works, for every such m.
If m>0: the sum is positive, so both roots, when real, are positive -- giving a real x, which is not what we want. We need the roots to be non-real instead, i.e.\ the discriminant strictly negative:
m2+16(a2−12)⟹a2<0<12−16m2.
Combining both branches, no real x exists exactly when
a2<12 if m≤0,ora2<12−16m2 if m>0.
For m≤0, the non-positive integers with a2<12 are
a∈{−3,−2,−1,0},independent of m.
For m>0, the same list shrinks as m grows:
At m=8: the bound becomes a2<8, giving only a∈{−2,−1,0}.
Once m≥83: the bound 12−m2/16 drops to 0 or below, so no non-positive integer a satisfies it at all.
Let α,β be the roots of x2−ax+b=0, and set Vn=αn+βn. Find a recurrence relating Vn+1, Vn, and Vn−1. Hence find the equation whose roots are αm and βm.
Understand the problem
Vn is just a name for `add the nth powers of both roots together' --- a new number for each n (so V1=α+β, V2=α2+β2, and so on). First we need to find a formula connecting three consecutive terms of this sequence. Then we use that same idea to build the equation whose two roots are αm and βm.
The question asks for a relation connecting Vn+1,Vn,Vn−1, and then the equation with roots αm,βm.
Computing power sums directly grows complicated fast. Instead, both roots satisfy the same quadratic relation, which expresses each higher power in terms of the two before it.
Multiply the defining relation by αn−1, and separately by βn−1. Add the two. This gives a recurrence for Vn in terms of a,b. Apply it repeatedly from V0,V1 up to Vm. Separately, the product αmβm needs none of that machinery: it's immediate from αβ.
Let t>0 be a root of ax2+bx+c=0. What can you say about the roots of 3ax2+2bx+c=0?
Understand the problem
We're given that t is a positive number solving the first equation. A second, related equation is built by tweaking the coefficients of the first. We need to figure out what can be said about where its roots sit, especially in relation to t.
The question asks what can be said about where the roots of 3ax2+2bx+c=0 lie.
The coefficients are arbitrary, so there's no clean closed form for the roots to inspect directly. Instead, use a connection to a cubic whose derivative is exactly this quadratic.
Construct a cubic whose derivative is exactly the new quadratic, by raising each of its terms one degree and dividing by the new exponent. This cubic vanishes at x=0, and, using that t solves the original quadratic, also at x=t. A function returning to the same value at two points has a stationary point strictly between them, so that stationary point of the cubic is exactly a root of the new quadratic, sitting strictly between 0 and t.
The target equation 3ax2+2bx+c=0 is exactly the derivative of ax3+bx2+cx, set to zero. This is what motivates looking at that cubic in the first place.
Let F(x)=ax3+bx2+cx. Then F(0)=0 always, and
F(t)=at3+bt2+ct=t(at2+bt+c)=t⋅0=0,
using the hypothesis that t is a root of ax2+bx+c=0. So F(0) and F(t) are both 0.
Since F is a polynomial, it is continuous and differentiable everywhere. Since F(0)=F(t) and t>0, F returns to the same value at both ends of [0,t], so its tangent must be flat somewhere strictly inside: there exists ξ∈(0,t) with F′(ξ)=0, as illustrated in Figure~\figref{D47rolle}.
Since F′(x)=3ax2+2bx+c, this says exactly that 3ax2+2bx+c=0 has a root at x=ξ∈(0,t).
Let (α,β) be the roots of x2−mx+A=0, and (γ,δ) be the roots of x2−nx+B=0 (m=n). Suppose (α,β,γ,δ) are in A.P. with α<β<γ<δ. Find (A,B) in terms of (m,n).
Understand the problem
`A.P.' (arithmetic progression) means each number is the one before it plus the same fixed amount every time, like 2,5,8,11 (adding 3 each step). We're told all four roots from these two separate equations, lined up in increasing order, form one single sequence like that. We need to find A and B, in terms of m and n, that make this true.
Two quadratics are given, each with an unknown constant term A,B, and sum-of-roots m,n respectively. The four roots together, in increasing order α<β<γ<δ, form an arithmetic progression, with α,β from the first equation and γ,δ from the second.
The question asks for every possible pair (A,B).
The ordering α<β<γ<δ does real work here: it pins down exactly which two of the four progression terms are the first equation's roots and which two are the second's, not just their sum and product.
Parametrize the four terms by a starting value t and step d. The ordering forces the first equation's roots to be the two smallest terms and the second's to be the two largest. Each equation's sum-of-roots relation then gives a linear equation in t,d. Solve, then each constant term follows as the product of its own pair.
Since α<β<γ<δ are exactly the four progression terms in increasing order, write
α=t,β=t+d,γ=t+2d,δ=t+3d,
for some starting value t and common difference d. Because (α,β) are the roots of the first quadratic and (γ,δ) the roots of the second, and α<β<γ<δ, the first quadratic's roots are exactly the two smallest terms and the second's are exactly the two largest --- no other pairing of the four terms is consistent with this ordering.
The sum-of-roots relation on the first quadratic gives α+β=m, i.e.
2t+d=m.
The sum-of-roots relation on the second quadratic gives γ+δ=n, i.e.
2t+5d=n.
Subtracting the first equation from the second: 4d=n−m, so
d=4n−m,t=85m−n.
Each constant term is the product of its own quadratic's pair of roots:
A=αβ=t(t+d)=6415m2+2mn−n2,B=γδ=(t+2d)(t+3d)=64−m2+2mn+15n2.
Suppose α,β satisfy: x2n+(2+3)nxn+(5)n=0 and βα and αβ satisfy xn+1+(x+1)n=0 for every integer n≥1. Prove that n is even.
Understand the problem
This is a `prove', not a `find' --- α,β satisfy a whole family of equations at once, one for every integer n≥1. We need to show this forces n to always be even. No computation needed, just a logical argument.
α,β are the roots of x2+cx+5=0, where c=2+3. Both also satisfy x2n+cnxn+(5)n=0, and α/β,β/α both satisfy xn+1+(x+1)n=0.
The question asks you to prove n is always even.
Writing α,β out explicitly is hopeless. Instead, each extra condition forces a relationship between αn+βn and c, and the two forced relationships are not obviously compatible.
Use the higher-degree equation, substituting Y=xn, to pin down αn+βn in terms of c. Use the ratio condition, substituting α/β directly and clearing denominators, to pin down αn+βn a second, independent way. Force the two expressions to agree.
Write c=2+3, so c=0. Since α,β are the roots of x2+cx+5=0: α+β=−c and αβ=5.
Because α,β also satisfy x2n+cnxn+(5)n=0, writing Y=xn shows that αn and βn are both roots of Y2+cnY+(5)n=0. Since αnβn=(αβ)n, and αβ=5 makes this (5)n, it automatically matches the product of roots of this quadratic, so the genuine constraint is on the sum:
αn+βn=−cn.
Separately, substituting x=α/β into xn+1+(x+1)n=0 gives
βnαn+1+(βα+β)n=0.
Multiplying through by βn:
αn+βn+(α+β)n=0⟹αn+βn=−(α+β)n⟹αn+βn=−(−c)n.
Combining the two expressions for αn+βn:
−cn⟹cn=−(−c)n=(−1)ncn.
Since c=2+3=0, this forces (−1)n=1, i.e.\ n is even.
Suppose x and m are real numbers related by
x2+(x+1)2x2=m2−1.
Find the number of real solutions x satisfying this equation, as a function of m.
Understand the problem
For a fixed real number m, this equation may have no real solution x at all, or several. For example, at x=1 the left side equals 1+41=45, so x=1 is a solution exactly when m2−1=45. We need to find, for every real m, how many real values of x satisfy the equation.
The question asks for the number of real values of x satisfying this equation, as a function of m.
Clearing the denominator gives a quartic in x --- don't try to solve that directly, it turns into a mess. Use the substitution below instead, which collapses it to a simple quadratic.
Let y=x+1x. Notice x−y equals xy. Naming p=xy, the original sum becomes a single quadratic in p, easy to solve in terms of m. Each solution for p then has to be checked separately for whether it produces real x.
Let y=x+1x (so x=−1). Then
x−y=x−x+1x=x+1x2,xy=x⋅x+1x=x+1x2,
so x−y=xy.
Write p:=xy, so p=x+1x2. Squaring x−y=p gives (x−y)2=p2, i.e.
x2−2xy+y2⟹(x2+y2)−2p=p2=p2.
Since y2=(x+1)2x2, this makes x2+y2 exactly the left side of the given equation, so x2+y2=m2−1 by hypothesis:
m2−1−2p=p2⟹p2+2p−(m2−1)=0⟹p=−1±1+(m2−1)⟹p=−1±∣m∣.
So p=m−1 or p=−m−1.
Each value of p came from p=x+1x2, i.e.\ x2−px−p=0, which has real roots exactly when its discriminant p(p+4) is non-negative, i.e.\ p≤−4 or p≥0.
Branch p=m−1: requiring p(p+4)≥0 gives
(m−1)(m+3)⟹m≥0≤−3 or m≥1.
Branch p=−m−1: requiring p(p+4)≥0 gives
(m+1)(m−3)⟹m≥0≤−1 or m≥3.
Every m≤−3 is also ≤−1, and every m≥3 is also ≥1, so at least one branch is active exactly when ∣m∣≥1. Hence no real x exists when ∣m∣<1.
For ∣m∣≥1, each active branch contributes x=2p±p(p+4). From branch p=m−1 (valid for m≤−3 or m≥1):
x=2(m−1)±(m−1)(m+3).
From branch p=−m−1 (valid for m≤−1 or m≥3):
x=2−(m+1)±(m+1)(m−3).
Counting the real values of x as ∣m∣ grows:
At ∣m∣=1: one of the two branches (whichever matches the sign of m) has its discriminant vanish exactly, giving the single repeated value x=0.
For 1<∣m∣<3: exactly one branch is active, giving two real values.
At ∣m∣=3: the other branch's discriminant also vanishes, giving a second repeated root, so the total becomes three distinct values.
For ∣m∣>3: both branches are fully active, giving four distinct values in total.
Suppose a,b,c are distinct and each pair of the equations x2+ax+bc=0, x2+bx+ca=0, x2+cx+ab=0 shares a common root. Find the roots.
Understand the problem
Three quadratics have coefficients that are each built from a,b,c, cycled into a different order each time. We need to find the three shared roots --- one for each pair of equations.
Eqn.~(1): x2+ax+bc=0. Eqn.~(2): x2+bx+ca=0. Eqn.~(3): x2+cx+ab=0, with a,b,c pairwise distinct. Every pair shares a root.
The question asks for each of the three pairwise common roots individually, and the condition on a,b,c under which this is possible.
Solving each quadratic in full is unwieldy. Instead: whenever two equations of the same shape share a root, subtracting one from the other eliminates the quadratic term, forcing a candidate value for the shared root directly.
Since a,b,c are pairwise distinct, subtracting Eqn.~(2) from Eqn.~(1) is a valid elimination step:
(a−b)x+(bc−ca)=0⟹(a−b)x=c(a−b)⟹x=c(since a=b).
So c is the shared root between Eqns.~(1) and (2).
By the same argument applied cyclically, x=a is forced for Eqns.~(2) and (3), and x=b for Eqns.~(1) and (3).
Checking that x=c genuinely satisfies Eqn.~(1):
c2+ac+bc=c(a+b+c)=0.
Excluding the trivial case c=0, this holds precisely when
a+b+c=0.