Question

if x^2+mx+m is a perfect square trinomial, which equation must be true

Answer 1

where are the choices?

Answer 2

hint:
\[(a+b)^2=a^2+2ab+b^2\]

Answer 3

One way to do this is to complete the square:
\[\begin{align*}x^2+mx+m&=x^2+mx+\frac{m^2}{4}-\frac{m^2}{4}+m\\
&=\left(x+\frac{m}{2}\right)^2-\frac{m^2}{4}+m\\
&=\left(x+\frac{m}{2}\right)^2+\frac{4m-m^2}{4}
\end{align*}\]
Since this is a perfect square trinomial, you must have
\[\frac{4m-m^2}{4}=0\]
Otherwise, you wouldn't be able to write \(x^2+mx+m\) as a squared binomial.