JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
Let $a,b\in\mathbb{C}$. Let $\alpha,\beta$ be the roots of the equation $x^{2}+ax+b=0$. If $\beta-\alpha=\sqrt{11}$ and $\beta^{2}-\alpha^{2}=3i\sqrt{11}$, then $(\beta^{3}-\alpha^{3})^{2}$ is equal to :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. Get $\alpha+\beta$ and $\alpha\beta$ from the data, then use the identity for $\beta^3-\alpha^3$. Formulas used. (1) $\beta^2-\alpha^2=(\beta-\alpha)(\beta+\alpha)$ and $(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta$(2) $\beta^3-\alpha^3=(\beta-\alpha)\big[(\alpha+\beta)^2-\alpha\beta\big]$Step 1 - Sum. By (1), $\alpha+\beta=\dfrac{3i\sqrt{11}}{\sqrt{11}}=3i$. Step 2 - Product. By (1), $11=(3i)^2-4\alpha\beta=-9-4\alpha\beta$, so $\alpha\beta=-5$. Step 3 - Cube difference. By (2), $$\beta^3-\alpha^3=\sqrt{11}\,(-9+5)=-4\sqrt{11}.$$ Step 4 - Value. $\left(\beta^3-\alpha^3\right)^2=16\cdot11=176$. Answer: (B)