JEE Main 2022 — Quadratic Equation Question with Solution

Given $\alpha ,\beta$ be the roots of the equation $x^2-\sqrt{2}x+\sqrt{6}=0$ 

So sum of roots will be $\alpha +\beta =\sqrt{2}$ and product of roots will be $\alpha \beta =\sqrt{6}$$$

And also given $\frac{1}{{\alpha }^2}+1$ and $\frac{1}{{\beta }^2}+1$ are roots of $x^2+ax+b=0$

So sum of roots will be $-a=\frac{1}{{\alpha }^2}+1+\frac{1}{{\beta }^2}+1$

$\Rightarrow a=\frac{-1}{{\alpha }^2}-\frac{1}{{\beta }^2}-2 .....\left(1\right)$

And similarly product of roots will be,

$b=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+1+\frac{1}{{\alpha }^2{\beta }^2} ....\left(2\right)$

Now adding equation $\left(1\right) \& \left(2\right)$ we get,

$a+b=\frac{1}{{\left(\alpha \beta \right)}^2}-1=\frac{1}{6}-1=-\frac{5}{6}$ {as $\alpha \beta =\sqrt{6}$}

Now putting the value of $a+b$ in $x^2-\left(a+b-2\right)x+\left(a+b+2\right)=0$

$\Rightarrow x^2-\left(-\frac{5}{6}-2\right)x+\left(2-\frac{5}{6}\right)=0$

$\Rightarrow 6x^2+17x+7=0$

$\Rightarrow x=-\frac{7}{3},x=-\frac{1}{2}$ are the roots, both roots are real and negative.