Question

if $x^2 + mx + m$ is a perfect - square trinomial, which equation must be true?\
\\(\bigcirc\\) $x^2 + mx + m=(x - 1)^2$\
\\(\bigcirc\\) $x^2 + mx + m=(x + 1)^2$\
\\(\bigcirc\\) $x^2 + mx + m=(x + 2)^2$\
\\(\bigcirc\\) $x^2 + mx + m=(x + 4)^2$

Explanation:

Step1: Recall perfect square form

A perfect square trinomial has the form $(x+a)^2 = x^2 + 2ax + a^2$.

Step2: Match coefficients with given trinomial

Given $x^2 + mx + m$, equate coefficients:

1. Coefficient of $x$: $m = 2a$
2. Constant term: $m = a^2$

Step3: Solve for $a$ and $m$

Substitute $m=2a$ into $m=a^2$:
$$2a = a^2$$
$$a^2 - 2a = 0$$
$$a(a-2)=0$$
Solutions: $a=0$ (invalid, since $m=0$ would make trinomial $x^2$) or $a=2$.
If $a=2$, then $m=2a=4$.

Step4: Verify the trinomial

Substitute $m=4$ into the given trinomial: $x^2 +4x +4=(x+2)^2$.

Answer:

C. $x^2 + mx + m = (x + 2)^2$