The quadratic function $f(x)=-x^2+3kx-2k^2+1$ has a maximum value of $(h^2-k)$, where $h$ and $k$ are constants.
By completing the square, I have found the maximum value of $f(x)$ in terms of $k$, that is $\frac{k^2}{4}+1$, which is correct.
However, the question requires me to show that $k=2h-2$, which I can, but I got another answer as well and am unsure why there are two answers. My working is as follows:
After completing the square, $f(x)=-(x-\frac{3}{2}k)^2+\frac{k^2}{4}+1$
When $x=\frac{3}{2}k$, maximum value $=h^2-k$
By equating the maximum value, $h^2-k=\frac{k^2}{4}+1$
$4h^2-4k=k^2+4$
$k^2+4k+4-4h^2=0$
$[k+(2+2h)][k+(2-2h)]=0$
Solving the equation, I get 2 answers:
$k+(2+2h)=0$
$k=-2h-2$
and
$k+(2-2h)=0$
$k=2h-2$ (shown)
Since substituting both answers satisfies the two expressions of the maximum value $h^2-k$ and $\frac{k^2}{4}+1$, may I know why and how to reject $k=-2h-2$, and why $k=2h-2$ is accepted?