Find the value of $\;$ $a$ $\;$ for which one root of the equation $\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;$ is twice as large as the other.
Given quadratic equation: $\;\;\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;\;\; \cdots \; (1)$
Given: $\;$ One root of equation $(1)$ is twice as large as the other.
Let the roots of $(1)$ be $\alpha, \; 2 \alpha$.
Sum of roots of equation $(1)$ $= \alpha + 2 \alpha = 3 \alpha = \dfrac{- \left(2a - 1\right)}{1} = 1 - 2a$
$\implies$ $\alpha = \dfrac{1 - 2a}{3}$ $\;\;\; \cdots \; (2)$
Product of roots of equation $(1)$ $= \alpha \times 2 \alpha = 2 \alpha^2 = \dfrac{a^2 + 2}{1}$
i.e. $\;$ $2 \alpha^2 = a^2 + 2$ $\;\;\; \cdots \; (3)$
In view of equation $(2)$, equation $(3)$ becomes
$2 \times \left(\dfrac{1 - 2a}{3}\right)^2 = a^2 + 2$
i.e. $\;$ $\dfrac{2 \left(1 - 4a + 4a^2\right)}{9} = a^2 + 2$
i.e. $\;$ $2 - 8a + 8a^2 = 9a^2 + 18$
i.e. $\;$ $a^2 + 8a + 16 = 0$
i.e. $\;$ $\left(a + 4\right)^2 = 0$
$\implies$ $a = -4$