# Math Help - quadratic equation

1. ## quadratic equation

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?

2. ## Re: quadratic equation

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.

4. ## Re: quadratic equation

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

5. ## Re: quadratic equation

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.

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

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

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

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

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

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

6. ## Re: quadratic equation

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

7. ## Re: quadratic equation

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!

8. ## Re: quadratic equation

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

9. ## Re: quadratic equation

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

10. ## Re: quadratic equation

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.

11. ## Re: quadratic equation

ok got it thank you both.

12. ## Re: quadratic equation

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.

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

13. ## Re: quadratic equation

Originally Posted by HallsofIvy
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 $\displaystyle a^2 = 9x^2 = (3x)^2$.

14. ## Re: quadratic equation

Thank you all again just one last thing to do with the discriminant.
if its equal to zero or a perfect square the quadratic is factorable is the correct?