Question

The value of m($k=$$m{a}^2$) for which the given value is a solution of the given equation :
$x^2+3ax+k=0,x=-a$ is

Answer 1

$x^2+3ax+k=0$

Substituting $x=-a$,

$\Rightarrow$ $(-a{)}^2+3a(-a)+k=0$

$\Rightarrow$ $a^2-3a^2+k=0$

$\Rightarrow$ $-2a^2+k=0$

$\Rightarrow$ $k=2a^2$

Now substituting $k=m{a}^2$

$\Rightarrow$ $m{a}^2=2a^2$

$\therefore$ $m=2$