Question

The value of $m$ for which one of the roots of $x^2-3x+2m=0$ is double of one of the roots of $x^2-x+m=0$, is

• A. $4$
• B. $-2$
• C. $2$
• D. $3$

Answer 1

The correct option is B $-2$

Let $\alpha$ be the roots of $x^2-x+m=0$ and $2\alpha$ be the root of $x^2-3x+2m=0$.

Then, ${\alpha }^2-\alpha +m=0$

and $4{\alpha }^2-6\alpha +2m=0$

On solving these equations, we get

$\frac{{\alpha }^2}{-2m+6m}=\frac{\alpha }{4m-2m}=\frac{1}{-6+4}$

$\Rightarrow \frac{{\alpha }^2}{4m}=\frac{\alpha }{2m}=\frac{1}{-2}$

$\Rightarrow {\alpha }^2=-2m$ and $\alpha =-m$

$\therefore (-m{)}^2=-2m$

$\Rightarrow m^2+2m=0$

$\Rightarrow m(m+2)=0$

$\Rightarrow m=0,m=-2$.