JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

Let $a, b \in C$ . Let $α, β$ be the roots of the equation $x^{2} + a x + b = 0$ . If $β - α = 11$ and $β^{2} - α^{2} = 3 i 11$ , then $(β^{3} - α^{3})^{2}$ is equal to :

A
160
B
176
C
194
D
187

Solution

Factor the second relation: $β^{2} - α^{2} = (β - α) (β + α) = 3 i 11$ . Since $β - α = 11$ , dividing gives $β + α = 3 i$ .

Then $α β = \frac{( β + α ) ^{2} - ( β - α ) ^{2}}{4} = \frac{( 3 i ) ^{2} - 11}{4} = \frac{- 9 - 11}{4} = - 5$ .

Use the identity $β^{3} - α^{3} = (β - α) ((β + α)^{2} - α β)$ , so

$(β^{3} - α^{3})^{2} = (β - α)^{2} ((β + α)^{2} - α β)^{2} = 11 \cdot (- 9 - (- 5))^{2} = 11 \cdot 16 = 176$ .

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