HI;
I need to complete the square for 9x^2 + 12x + 4.
but I can't factor out the 9.

therefore how do I start?

Originally Posted by anthonye
HI;
I need to complete the square for 9x^2 + 12x + 4.
but I can't factor out the 9.

therefore how do I start?
no one said coefficients have to be integers

\begin{align*}&9x^2+12x+4= \\ \\ &9\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)= \\ \\ &9\left(x+\frac{2}{3}\right)^2= \\ \\ &\left(3x+2\right)^2 \end{align*}

take your pick of the last two forms

Ok thanks.

I just need confirmation that the following steps are correct?

9x2 +12x + 4 = 0
9x2 + 12x = - 4
9(x2 + 4/3 + 4/9) = - 4
9(x + 2/3)(x + 2/3) = 0

9(x + 2/3)2 = 0
(x + 2/3) = 0
X = + or – 2/3
= 0.66667

Originally Posted by anthonye
I just need confirmation that the following steps are correct?

9x2 +12x + 4 = 0
9x2 + 12x = - 4
9(x2 + 4/3 + 4/9) = - 4
9(x + 2/3)(x + 2/3) = 0

9(x + 2/3)2 = 0
(x + 2/3) = 0
X = + or – 2/3
= 0.66667
No.

$$9x^2+12x+4=0$$

$$9\left(x^2+\frac{12}{9}+\frac{4}{9}\right)=0$$

$$9\left(x^2+\frac{4}{3}+\frac{4}{9}\right)=0$$

$$9\left(\left(x+\frac{2}{3}\right)^2\right)=0$$

$$\left(x+\frac{2}{3}\right)^2=0$$

$$x=-\frac{2}{3}$$

9x^2 +12x + 4 = 0
9x^2 + 12x = - 4
9(x^2 + 4/3x + 4/9) = - 4
9(x + 2/3)(x + 2/3) = 0

9(x + 2/3)^2 = 0
(x + 2/3) = 0
X = + or – 2/3
= 0.66667

Anthonye, did you not notice from Romsek's first post that you do NOT "need to complete the square"? $9x^2+ 12x+4= (3x+2)^2$ is already a perfect square!

Thanks I didn't notice that.also mostly I just need to know if the process is correct?

Also how can you tell if a quadratic is a perfect square?

You start by learning what a "perfect square" is! For any numbers, a and b, $(a+ b)^2=a^2+ 2ab+ b^2$. The quadratic here is $9x^2+ 12x+ 4$. Clearly, $a^2= x^2$ and $b^2= 4$ so a= 3x and b= 2. From that 2ab= 2(3x)(2)= 12x. The whole point of "completing the square" is adding and subtracting that "2ab" part. I don't see how you can learn "completing the square" without first knowing what you want to get.

ok got it thank you both.

Originally Posted by anthonye
ok got it thank you both.
I want to make sure you understand that x=2/3 is NOT a root.

$$\left(x+\frac{2}{3}\right)^2 \left|_{x=\frac{2}{3}}\right.=\frac{16}{9} \neq 0$$

You start by learning what a "perfect square" is! For any numbers, a and b, $(a+ b)^2=a^2+ 2ab+ b^2$. The quadratic here is $9x^2+ 12x+ 4$. Clearly, $a^2= x^2$ and $b^2= 4$ so a= 3x and b= 2. From that 2ab= 2(3x)(2)= 12x. The whole point of "completing the square" is adding and subtracting that "2ab" part. I don't see how you can learn "completing the square" without first knowing what you want to get.
Actually $\displaystyle a^2 = 9x^2 = (3x)^2$.