# How do you solve 3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2?

Jun 16, 2016

Real solution: $x = 1$

Complex solutions: $x \in \left\{1 , \sqrt{2} i , - \sqrt{2} i\right\}$

#### Explanation:

L.S. $= 3 {x}^{3} - 2 \left[x - \left(3 - 2 x\right)\right]$

$\textcolor{w h i t e}{\text{XXX}} = 3 {x}^{3} - 2 \left[- 3 + 3 x\right]$

$\textcolor{w h i t e}{\text{XXX}} = 3 {x}^{3} - 6 x + 6$

R.S. $= 3 {\left(x - 2\right)}^{2}$

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

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

Since L.S. $=$ R.S.
$3 {x}^{3} + 6 x + 6 = 3 {x}^{2} - 12 x + 12$

$\rightarrow {x}^{3} + 2 x + 2 = {x}^{2} - 4 x + 4$

$\rightarrow \textcolor{g r e e n}{1} {x}^{3} - {x}^{2} + 2 x - \textcolor{b l u e}{2} = 0$

In the hope of finding rational roots we try the factors of $\frac{\textcolor{b l u e}{2}}{\textcolor{g r e e n}{1}}$

and we find $\textcolor{red}{x = 1}$ is a root
$\rightarrow \left(x - 1\right)$ is a factor.

$\left({x}^{3} - {x}^{2} + 2 x - 2\right) \div \left(x - 1\right) = {x}^{2} + 2$

$\left({x}^{2} + 2\right)$ has no Real roots (i.e. no Real solutions for $x$)

However, if we allow Complex solutions:
$\textcolor{w h i t e}{\text{XXX}} x = \pm \sqrt{2} i$