# Completing the square #2

#### Mukilab

$$\displaystyle 9x^2+12x+7$$

Can I do

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

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$$\displaystyle 9x^2+12x+7$$

Can I do

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

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

Alternatively,

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

#### sa-ri-ga-ma

$$\displaystyle 9x^2+12x+7$$

Can I do

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

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

#### Prove It

MHF Helper
$$\displaystyle 9x^2+12x+7$$

Can I do

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

So what you need to do is...

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

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

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

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

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

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