Completing the square #2

Nov 2009
468
14
\(\displaystyle 9x^2+12x+7\)

Can I do

\(\displaystyle (9x+12x+36)-29\)?
 
Last edited by a moderator:
Dec 2009
3,120
1,342
\(\displaystyle 9x^2+12x+7\)

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
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle 9x^2+12x+7\)

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...

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

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

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

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

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

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