JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^{2}+\lambda x-1=0$ is $15$, then $6(\alpha^{3}+\beta^{3})^{2}$ is equal to :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. The condition gives $\lambda^2$; then compute $\alpha^3+\beta^3$. Formulas used. (1) $\frac1{\alpha^2}+\frac1{\beta^2}=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$ and $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$Step 1 - Find $\lambda^2$. $\alpha+\beta=-\frac\lambda3$, $\alpha\beta=-\frac13$. By (1), $$9\left(\frac{\lambda^2}{9}+\frac23\right)=\lambda^2+6=15\ \Longrightarrow\ \lambda^2=9.$$ Step 2 - Cubes. By (1), $$\alpha^3+\beta^3=-\frac{\lambda^3}{27}-\frac\lambda3=\mp2\quad(\lambda=\pm3).$$ Step 3 - Value. $6\cdot(\pm2)^2=24$. Answer: (B)