Quadratic Equations (JEE Main)

Question 1

Let α and β be the roots of x²−2x+4=0. Find the value of (α¹⁰+β¹⁰)/(α⁴+β⁴).

Accepted Answer

64. Key insight: the roots are 1± i√3=2e^(± iπ/3), so αⁿ+βⁿ=2· 2ⁿcos(nπ)/3.
Numerator: 2· 2¹⁰cos(10π)/3=2·1024·(−1/2)=−1024.
Denominator: 2· 2⁴cos(4π)/3=2·16·(−1/2)=−16.
Ratio =(−1024)/(−16)=64.
Trap: grinding the recurrence Sₙ₊₂=2Sₙ₊₁−4Sₙ up to n=10 invites arithmetic slips; students who drop the signs of the cosines still get 64 by luck, but those who take cos(10π)/3=+1/2 get −64.
Exam pattern: JEE

Question 2

Let k be a complex number such that one root of x²−kx+8=0 is the square of the other root. Find the product of all possible values of k.

Accepted Answer

72. Let the roots be t and t². Then t· t²=8, so t³=8 and t∈{2, 2ω, 2ω²}.
k=t+t² gives k₁=2+4=6, k₂=2ω+4ω², k₃=2ω²+4ω.
k₂k₃=4ω³+8ω²+8ω⁴+16ω³=20+8(ω+ω²)=20−8=12.
Product =6·12=72.
Trap: taking only the real cube root gives the single value k=6 and answer 6; the word complex makes all three cube roots admissible.
Exam pattern: JEE Main – one root is a power of the other, cube roots of unity

Question 3

For a real parameter a, the equation x²−2ax+2a²−4a+3=0 has real roots α,β. If M and m are the greatest and least values of α²+β², find M+m.

Accepted Answer

20. Reality: D/4=a²−(2a²−4a+3)=−(a−1)(a−3)≥0, so 1≤ a≤3.
α²+β²=(2a)²−2(2a²−4a+3)=8a−6, which is linear and increasing in a.
On [1,3]: m=8−6=2 (roots 1,1) and M=24−6=18 (roots 3,3).
M+m=20.
Trap: ignoring the reality condition makes 8a−6 unbounded, so no extremes exist; the hidden constraint 1≤ a≤3 is the whole question.
Exam pattern: JEE Main – extreme value of α²+β² with discriminant restriction on t

Quadratic Equations: questions with solutions

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