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

Correct Answer: 98

∵ 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

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

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

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