# Math Help - Completing the square #2

1. ## Completing the square #2

$9x^2+12x+7$

Can I do

$(9x+12x+36)-29$?

2. Originally Posted by Mukilab
$9x^2+12x+7$

Can I do

$(9x\color{red}^2\color{black}+12x+36)-29$?
Not quite,

$9x^2+12x+7=(3x)^2+12x+7=(3x)^2+2(6x)+7=$ $(3x)^2+2[3(2x)]+7=(3x)^2+2[3(2x)]+2^2-2^2+7=(3x+2)^2+7-4$

Alternatively,

$9\left(x^2+\frac{12}{9}x\right)+7=9\left(x^2+2\lef t[\frac{6}{9}x\right]+\left[\frac{6}{9}\right]^2\right)+7-9\left(\frac{6}{9}\right)^2$ etc

3. Originally Posted by Mukilab
$9x^2+12x+7$

Can I do

$(9x+12x+36)-29$?
$9x^2 + 2*3*2*x + 4 + 3$

$(3x + 2)^2 + 3$

4. Originally Posted by Mukilab
$9x^2+12x+7$

Can I do

$(9x+12x+36)-29$?
No, the coefficient of $x^2$ needs to be $1$.

So what you need to do is...

$9x^2 + 12x + 7 = 9\left(x^2 + \frac{4}{3}x + \frac{7}{9}\right)$

$= 9\left[x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \frac{7}{9}\right]$

$= 9\left[\left(x + \frac{2}{3}\right)^2 - \frac{4}{9} + \frac{7}{9}\right]$

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

$= 9\left[\left(x + \frac{2}{3}\right)^2 + \frac{1}{3}\right]$

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