Question

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

X^2+mx
X^2+mx+m(x+2)^2
X^2+mx+m(x+2)^2
X^2+mx+m(x+4)^2

Answer 1

For the first one it is x^2+mx+m=(x-1)^2

Answer 2

You might want to practice "completing the square" with a specific problem, and then go back and follow the same steps when only 'm' the variable appears.

How would we "complete the square" if given x^2 + 6x?

1) Take half of the coefficient of x, and then square your result: (1/2)(6)=3 => 3^2 = 9

2) Add this square (9) and then subtract this square (9):

x^2 + 6x + 9 - 9

3) Rewrite x^2 + 6x + 9 as the square of a binomial: (x+3)^2

4) Rewrite the original expression, x^2 + 6x, as (x+3)^2 - 9

Follow these steps with x^2 + mx + m. Can you find a value for m that makes
x^2 + mx + m a "perfect square trinomial?"

Answer 3

\[\left( \frac{ m }{ 2 } \right)^2=m,m^2=4m,m^2-4m=0,m(m-4)=0,m=0,\]
because then it is not a trinomial,
then m=4