Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - 4\lambda x + 5 = 0$ and α, γ 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

∵ α, β are roots of $x^2 - 4\lambda x + 5 = 0$
$α + β = 4λ$ and $αβ = 5$

Also, α, γ are roots of
$x^2 - (3\sqrt{2} + 2\sqrt{3})x + 7 + 3\sqrt{3}\lambda = 0, \quad \lambda > 0$
∴$$\alpha + \gamma = 3\sqrt{2} + 2\sqrt{3}, $\alpha\gamma = 7 + 3\sqrt{3}\lambda$
$∵$ $α$ is common root

$∴$ \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

$\alpha = \frac{2 + 3\sqrt{3}\lambda}{3\sqrt{2} + 2\sqrt{3} - 4\lambda}$
$∵$ $\beta + \gamma = 3\sqrt{2}$
∴$$4\lambda + 3\sqrt{2} + 2\sqrt{3} - 2\alpha = 3\sqrt{2}

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

$∴$ $\lambda = \frac{6\sqrt{2} - 3\sqrt{3} \pm \sqrt{9(11-4\sqrt{6}) + 32(4+3\sqrt{6})}}{16}$
$∴$ $λ=2$
$∴$ $(\alpha + 2\beta + \gamma)^2 = (\alpha + \beta + \beta + \gamma)^2 = (4\sqrt{2} + 3\sqrt{2})^2 = (7\sqrt{2})^2 = 98$