JEE Main 2022 — Mathematics Algebra

Given \alpha &\beta are roots of ${x}^{2}-4\lambda x+5=0$x2−4λx+5=0

So, $\alpha +\beta =4\lambda$α+β=4λ and $\alpha \beta =5$αβ=5 .......(i)

And \alpha &\gamma are roots of ${x}^{2}-(3\sqrt{2}+2\sqrt{3})x+(7+3\lambda \sqrt{3})=0$x2−(32​+23​)x+(7+3λ3​)=0

So, $\alpha +y=3\sqrt{2}+2\sqrt{3}$α+y=32​+23​ and $\alpha \gamma =7+3\lambda \sqrt{3}$αγ=7+3λ3​ .........(ii)

Given, if $\beta +\gamma =3\sqrt{2}$β+γ=32​

So, from equation (i) and (ii) we get, $\alpha =2\lambda +\sqrt{3}$α=2λ+3​ and

$\beta =2\lambda -\sqrt{3}$β=2λ−3​, Now by product of roots we get, $4{\lambda }^{2}-3=5\Rightarrow \lambda =\sqrt{2}$4λ2−3=5⇒λ=2​

$$\therefore {(\alpha +2\beta +\lambda )}^{2}={(\alpha +\beta +\beta +\lambda )}^{2}={(4\lambda +3\sqrt{2})}^{2}$$∴(α+2β+λ)2=(α+β+β+λ)2=(4λ+32​)2

$$={(4\sqrt{2}+3\sqrt{2})}^{2}={(7\sqrt{2})}^{2}=98$$=(42​+32​)2=(72​)2=98