Quadratic Equations and Root Relations (JEE Advanced)

Question 1

Let α,β be the roots of x²+4x+2=0. The quadratic equation whose roots are (α²)/β and (β²)/α is underline. Exactly one option is correct.

Accepted Answer

B. Here α+β=−4 and αβ=2. Sum of new roots: (α²)/β+(β²)/α=(α³+β³)/(αβ), and α³+β³=(α+β)³−3αβ(α+β)=−64+24=−40, so the sum is −20. Product of new roots: (α²)/β·(β²)/α=αβ=2. Hence the equation is x²−(−20)x+2=0, i.e. x²+20x+2=0. (Using (α+β)³/αβ=−32 for the sum, i.e. forgetting the −3αβ(α+β) term, gives the distractor x²+32x+2=0.)
Difficulty: Hard

Question 2

Let a,b,c be real numbers with a≠0, and let f(x)=ax²+bx+c. Which of the following conditions, on its own, guarantees that the equation f(x)=0 has real roots? One or more options may be correct.

Accepted Answer

B, C, D. Key fact: if a real quadratic takes values of opposite signs at two points, or its value at some point has sign opposite to the leading coefficient, then its graph crosses the x-axis, so the roots are real (and distinct).
(A) c(a+b+c)>0 means f(0)f(1)>0: no sign change is forced. Example a=1,b=0,c=1: c(a+b+c)=2>0 but x²+1=0 has no real root. Not sufficient.
(B) a(a+b+c)=a f(1)<0. Then af(x)=a²x²+a

Question 3

Real numbers p,q,r are such that the equation 14x²−2(p+2q+3r)x+(p²+q²+r²)=0 has a real root, and p+q+r=12. Find the value of pqr. Give the answer as an integer.

Accepted Answer

48. Observe that 14x²−2(p+2q+3r)x+p²+q²+r²=(x−p)²+(2x−q)²+(3x−r)² (expand: x²+4x²+9x²=14x², cross terms −2px−4qx−6rx). For a real root x, a sum of three real squares is zero, so p=x, q=2x, r=3x. Then p+q+r=6x=12 gives x=2, so (p,q,r)=(2,4,6) and pqr=48. (Equivalently, the discriminant is 4[(p+2q+3r)²−14(p²+q²+r²)]≤0 by the Cauchy-Schwarz inequality, so a real root forces equality, i.e. p:q:r=1:2:3.)
D

Quadratic Equations and Root Relations: questions with solutions

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