QJEEM22Ma9e80: JEE Main 2022 Mathematics - Class 11 JEEM22Ma9e80 Solution: Let , be the roots of the equation and , be the roots of the…

$\because$ $\alpha$ , $\beta$ are roots of x² $-$ 4 $\lambda$ x + 5 = 0

$\therefore$ $\alpha$ + $\beta$ = 4 $\lambda$ and $\alpha$ $\beta$ = 5

Also, $\alpha$ , $\gamma$ are roots of

{x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0

$\therefore$ $\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3$ , $\alpha \gamma = 7 + 3\sqrt 3 \lambda$

$\because$ $\alpha$ is common root

$\therefore$ ${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$ ....... (i)

and ${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$ ...... (ii)

From (i) - (ii) : we get $\alpha = {{2 + 3\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$

$\because$ $\beta + \gamma = 3\sqrt 2$

$\therefore$ $4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2$

$\Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$

$\Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$

$\therefore$ $\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$

$\therefore$ $\lambda = \sqrt 2$

$\therefore$ ${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$

$= {(4\sqrt 2 + 3\sqrt 2 )^2}$

$= {(7\sqrt 2 )^2} = 98$