# How do you solve 9x² - 12x + 4 = -3?

Nov 8, 2017

$x = \frac{2 \pm \sqrt{3} i}{3}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 9 {x}^{2} - 12 x + 4 = - 3$

There are several ways to solve this.
I will demonstrate using a "completing the square method"

The given equation implies
$\textcolor{w h i t e}{\text{XXX}} 9 {x}^{2} - 12 x = - 7$ (after subtracting $4$ from both sides)

$\textcolor{w h i t e}{\text{XXX}} 9 \left({x}^{2} - \frac{4}{3} x\right) = - 7$

$\textcolor{w h i t e}{\text{XXX}} 9 \left({x}^{2} - \frac{4}{3} x + {\left(\frac{2}{3}\right)}^{2}\right) = - 7 + 9 \cdot {\left(\frac{2}{3}\right)}^{2}$

$\textcolor{w h i t e}{\text{XXX}} 9 {\left(x - \frac{2}{3}\right)}^{2} = - 7 + 4$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - \frac{2}{3}\right)}^{2} = - \frac{3}{9} = - \frac{1}{3}$

$\textcolor{w h i t e}{\text{XXX}} \left(x - \frac{2}{3}\right) = \pm \sqrt{- \frac{1}{3}}$

Note that for Real values $\sqrt{- \frac{1}{3}}$ is undefined,
but if we are allowed Complex values:
$\textcolor{w h i t e}{\text{XXX}} x - \frac{2}{3} = \pm \frac{1}{\sqrt{3}} i$
and
$\textcolor{w h i t e}{\text{XXX}} x = \frac{2}{3} \pm \frac{1}{\sqrt{3}} i = \frac{2 \pm \sqrt{3} i}{3}$

Why is there not Real solution?
Note that the given equation is equivalent to $9 {x}^{2} - 12 x + 7 = 0$
and here is a graph of $9 {x}^{2} - 12 x + 7$
graph{9x^2-12x+7 [-12.41, 12.9, -4.42, 8.22]}
Notice that $9 {x}^{2} - 12 x + 7$ never crosses the X-axis and therefore it is never equal to $0$