JEE Main 2022 — Mathematics Algebra

Given $\alpha ,\beta$α,β be the roots of the equation ${x}^{2}-\sqrt{2}x+\sqrt{6}=0$x2−2​x+6​=0

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

And also given $\frac{1}{{\alpha }^{2}}+1$α21​+1 and $\frac{1}{{\beta }^{2}}+1$β21​+1 are roots of ${x}^{2}+ax+b=0$x2+ax+b=0

So sum of roots will be $-a=\frac{1}{{\alpha }^{2}}+1+\frac{1}{{\beta }^{2}}+1$−a=α21​+1+β21​+1

$$\Rightarrow a=\frac{-1}{{\alpha }^{2}}-\frac{1}{{\beta }^{2}}-2.....(1)$$⇒a=α2−1​−β21​−2.....(1)

And similarly product of roots will be,

$$b=\frac{1}{{\alpha }^{2}}+\frac{1}{{\beta }^{2}}+1+\frac{1}{{\alpha }^{2}{\beta }^{2}}....(2)$$b=α21​+β21​+1+α2β21​....(2)

Now adding equation (1)&(2) we get,

$a+b=\frac{1}{{(\alpha \beta )}^{2}}-1=\frac{1}{6}-1=-\frac{5}{6}$a+b=(αβ)21​−1=61​−1=−65​ {as $\alpha \beta =\sqrt{6}$αβ=6​}

Now putting the value of $a+b$a+b in ${x}^{2}-(a+b-2)x+(a+b+2)=0$x2−(a+b−2)x+(a+b+2)=0

$$\Rightarrow {x}^{2}-(-\frac{5}{6}-2)x+(2-\frac{5}{6})=0$$⇒x2−(−65​−2)x+(2−65​)=0

$$\Rightarrow 6{x}^{2}+17x+7=0$$⇒6x2+17x+7=0

$\Rightarrow x=-\frac{7}{3},x=-\frac{1}{2}$⇒x=−37​,x=−21​ are the roots, both roots are real and negative.