Question

Given two quadratic equations $x^2-x+m=0$ and $x^2-x+3m=0,m\ne 0.$ 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.

Answer 1

Let $\alpha$ be the root of first equation

$\therefore 2\alpha$ is the root of second equation

${\alpha }^2-\alpha +m=0$ $\to \left(1\right)$

$4{\alpha }^2-2\alpha +3m=0$ $\to \left(2\right)$

$4{\alpha }^2-2\alpha +3(\alpha -{\alpha }^2)=0$

${\alpha }^2+\alpha =0$

$\alpha (\alpha +1)=0$

$\alpha =0$ or $\alpha =-1$

Put $\alpha =0$ i equation (1)

$0-0+m=0$

$m=0$

Put $\alpha =-1$ in equation (1)

$1+1+m=0$

$m=-2$

$m=-2,0$