In each of the following, find the value of $k$ for which the given value is a solution of the given equation:
$x^2+3ax+k=0$, $x=-a$
Given:
Given equation is $x^2+3ax+k=0$.
To do:
We have to find the value of $k$ for which $x=-a$ is a solution of $x^2+3ax+k=0$.
Solution:
If $x=m$ is a solution of $f(x)$ then $f(m)=0$.
Therefore,
For $x=-a$
$x^2+3ax+k=0$
$(-a)^2+3a(-a)+k=0$
$a^2-3a^2+k=0$
$-2a^2+k=0$
$k=2a^2$
The value of $k$ is $2a^2$.
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