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

Answer: 98

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