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Mathematics · Quadratic Equations

JEE Main 2026 — 5 April, Morning Shift — Question 24

Let a,b∈C.a, b ∈ C. Let α,βα, β be the roots of the equation x2+ax+b=0x² + ax + b = 0. If β−α=11β - α = \sqrt11 and β2−α2=3i11β² - α² = 3i\sqrt{11}, then (β3−α3)2(β³ - α³)² is equal to :

  1. Option A:

    160

  2. Option B:

    176

    Correct
  3. Option C:

    194

  4. Option D:

    187

Answer: B

Step-by-step solution

β−α=11\beta-\alpha=\sqrt{11} and β2−α2=3i11\beta^{2}-\alpha^{2}=3 \mathrm{i} \sqrt{11} ⇒(β+α)(β−α)=3i11\Rightarrow(\beta+\alpha)(\beta-\alpha)=3 \mathrm{i} \sqrt{11}

\Rightarrow \beta+\alpha=3 \mathrm{i} \end{gathered}$$ from (1) and (2) $\beta=\frac{\sqrt{11}+3 \mathrm{i}}{2}, \alpha=\frac{3 \mathrm{i}-\sqrt{11}}{2}$ $\Rightarrow \alpha \beta=\frac{-9-11}{4} \Rightarrow \alpha \beta=-5$ Now $\left(\beta^{3}-\alpha^{3}\right)^{2}=(\beta-\alpha)^{2}\left(\beta^{2}+\alpha^{2}+\alpha \beta\right)^{2}$ $=(\beta-\alpha)^{2}\left((\beta+\alpha)^{2}-\alpha \beta\right)^{2}$ $=(11)(-9+5)^{2}=176$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Mathematics
Chapter
Quadratic Equations
Topic
Theory of Quadratic Equations

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