Question

Let $\alpha, \beta$ be the roots of the equation $x^{2}-4 \lambda x+5=0$ and $\alpha, \gamma$ be the roots of the equation $x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+7+3 \lambda \sqrt{3}=0$ If $\beta+\gamma=3 \sqrt{2}$, then $(\alpha+2 \beta+\gamma)^{2}$ is equal to:

Answer 1

$x^{2}-4 \lambda x+5=0\left\langle_{\beta}^{\alpha}\right.$
$x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+(7+3 \lambda \sqrt{3})=0\left\langle_{\gamma}^{\alpha}\right.$
$\alpha+\beta=4 \lambda$
$\alpha+\gamma=3 \sqrt{2}+2 \sqrt{3}$
$\beta+\lambda=3 \sqrt{2}$
$\therefore \alpha=2 \lambda+\sqrt{3}$
$\beta=2 \lambda-\sqrt{3}$
$\alpha \gamma=7+3 \lambda \sqrt{3}$
$\alpha \beta=5$
$4 \lambda^{2}=8 \Rightarrow \lambda=\sqrt{2}$
$\therefore (\alpha+2 \beta+\lambda)^{2}=(4 \alpha+3 \sqrt{2})^{2}=(7 \sqrt{2})^{2}=98$