Given two quadratic equations $x^2 - x + m = 0$ and $x^2 - x + 3m = 0$, $m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is equal to double the root of the first equation.
Given quadratic equations: $\;$ $x^2 - x + m = 0$ $\;\;\; \cdots \; (1)$
$x^2 - x + 3m = 0$ $\;\;\; \cdots \; (2)$
Let $\alpha_1$ and $\beta_1$ be the roots of equation $(1)$.
Let $\alpha_2$ and $\beta_2$ be the roots of equation $(2)$.
As per question, $\;$ $\alpha_2 = 2 \alpha_1$ $\;\;\; \cdots \; (3)$
Now, $\;$ $\alpha_1 + \beta_1 = -1$ $\;\;\; \cdots \; (4a)$; $\;\;$ $\alpha_1 \cdot \beta_1 = m$ $\;\;\; \cdots \; (4b)$
$\alpha_2 + \beta_2 = -1$ $\;\;$ i.e. $\;\;$ $2 \alpha_1 + \beta_2 = -14$ $\;\;\; \cdots \; (5a)$ $\;\;\;$ [in view of (3)]
and $\;$ $\alpha_2 \cdot \beta_2 = 3m$ $\;\;$ i.e. $\;\;$ $2 \alpha_1 \cdot \beta_2 = 3m$ $\;\;\; \cdots \; (5b)$ $\;\;\;$ [in view of (3)]
We have from equations $(4a)$ and $(5a)$,
$2 \beta_1 - \beta_2 = -1$ $\;\;\; \cdots \; (6a)$
From equation $(4b)$, $\;$ $\alpha_1 = \dfrac{m}{\beta_1}$
From equation $(5b)$, $\;$ $\alpha_1 = \dfrac{3m}{2 \beta_2}$
$\therefore \;$ We have, $\;$ $\dfrac{m}{\beta_1} = \dfrac{3m}{2 \beta_2}$
i.e. $\;$ $\beta_2 = \dfrac{3}{2} \beta_1$ $\;\;\; \cdots \; (6b)$ $\;\;$ provided $\;$ $m \neq 0$
In view of equation $(6b)$, equation $(6a)$ becomes
$2 \beta_1 - \dfrac{3}{2} \beta_1 = -1$
i.e. $\;$ $\beta_1 = -2$
Substituting the value of $\beta_1$ in equation $(4b)$ gives
$\alpha_1 = -1 - \beta_1 = -1 + 2 = 1$
Substituting the values of $\alpha_1$ and $\beta_1$ in equation $(4b)$ gives
$1 \times \left(-2\right) = m$ $\implies$ $m = -2$