# How do you solve using the completing the square method 9x^2-12x+5=0?

Mar 7, 2016

I found:
${x}_{1} = \frac{2 + i}{3}$
${x}_{2} = \frac{2 - i}{3}$

#### Explanation:

Let us manipulate a bit our expression (take the $5$ to the right):
$9 {x}^{2} - 12 x = - 5$
let us add and subtract $4$ to the left:
$9 {x}^{2} - 12 x \textcolor{red}{+ 4 - 4} = - 5$
rearrange:
$9 {x}^{2} - 12 x + 4 = 4 - 5$
let us recognize the square on the left as:
$\textcolor{b l u e}{{\left(3 x - 2\right)}^{2}} = - 1$
BUT
if we try to solve taking the root of both sides we will get a negative square root!
I am not sure you know about them but we can write $\sqrt{- 1} = i$ (the immaginarty unit) and keep on going solving our equation as:
$\sqrt{{\left(3 x - 2\right)}^{2}} = \sqrt{- 1}$
$3 x - 2 = \pm i$
so we have 2 solutions:
${x}_{1} = \frac{2 + i}{3}$
${x}_{2} = \frac{2 - i}{3}$