# How do you solve by completing the square (x -2) (x + 3)= x - 10?

Mar 6, 2018

$x = \pm 2 i$

#### Explanation:

Given: $\left(x - 2\right) \left(x + 3\right) = x - 10$

$\textcolor{b r o w n}{\text{Consider just the left hand side (LHS)}}$

$\textcolor{b l u e}{\left(x - 2\right)} \textcolor{g r e e n}{\left(x + 3\right)}$

Multiply everything in the right brackets by everything in the left.

$\textcolor{g r e e n}{\textcolor{b l u e}{x} \left(x + 3\right) \textcolor{w h i t e}{\text{ddd}} \textcolor{b l u e}{- 2} \left(x + 3\right)} \leftarrow$ Notice the minus followed the 2

$\textcolor{g r e e n}{{x}^{2} + \underbrace{3 x \textcolor{w h i t e}{\text{dddd}} - 2 x} - 6}$

$\textcolor{g r e e n}{{x}^{2} \textcolor{w h i t e}{\text{dddddd")+xcolor(white)("ddd}} - 6}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Putting it all back together}}$

color(white)("dd")"LHS"color(white)("ddddd")=color(white)("ddd")"RHS"
${x}^{2} + x - 6 \textcolor{w h i t e}{\text{dd")=color(white)("dd}} x - 10$

As $x$ is on both sides we can cancel them out.

${x}^{2} + \cancel{x} - 6 \textcolor{w h i t e}{\text{dd")=color(white)("dd}} \cancel{x} - 10$

${x}^{2} + 4 = 0$
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$\textcolor{b r o w n}{\text{Completing the square}}$ ( If they insist !!!)

Write as ${x}^{2} + 0 x + 4 = 0$

${\left(x + \frac{0}{2}\right)}^{2} + 4 = 0$

${\left(x + \frac{0}{2}\right)}^{2} = - 4$

Square root both sides

$x + \frac{0}{2} = \pm \sqrt{- 4}$

$x = \pm \sqrt{4 \times \left(- 1\right)}$

$x = \pm 2 i$

Mar 6, 2018

This solution is irrational. See below...

Hope this helps!