JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
Let $\alpha,\beta$ be the roots of the equation $x^2-\sqrt2x+\sqrt6=0$ and $\frac1{\alpha^2}+1,\frac1{\beta^2}+1$ be the roots of the equation $x^2+ax+b=0$. Then the roots of the equation $x^2-(a+b-2)x+(a+b+2)=0$ are :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. Find $a$ and $b$ from the new roots, then check the sign of the discriminant, sum and product of the last equation. Formulas used. (1) $\frac1{\alpha^2}+\frac1{\beta^2}=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$Step 1 - Reciprocal squares. $\alpha+\beta=\sqrt2$, $\alpha\beta=\sqrt6$. By (1), $$\frac1{\alpha^2}+\frac1{\beta^2}=\frac{2-2\sqrt6}{6}=\frac{1-\sqrt6}{3}.$$ Step 2 - $a$ and $b$. $$-a=\frac{1-\sqrt6}{3}+2=\frac{7-\sqrt6}{3},\qquad b=\frac16+\frac{1-\sqrt6}{3}+1=\frac{9-2\sqrt6}{6}.$$ So $a+b=\dfrac{-14+2\sqrt6+9-2\sqrt6}{6}=-\dfrac56$. Step 3 - Last equation. $x^2+\frac{17}{6}x+\frac76=0$. Discriminant $\frac{289-168}{36}>0$, sum $<0$, product $>0$: both roots real and negative. Answer: (B)