# How do you simplify x^2+x+2=0 ?

Feb 1, 2017

This will only "simplify" into linear factors with the help of Complex coefficients as:

$\left(x + \frac{1}{2} - \frac{\sqrt{7}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{7}}{2} i\right) = 0$

#### Explanation:

Given:

${x}^{2} + x + 2 = 0$

Note that this is in the form:

$a {x}^{2} + b x + c = 0$

with $a = 1$, $b = 1$ and $c = 2$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{2}\right) = 1 - 8 = - 7$

Since $\Delta < 0$ this quadratic has no simpler factors with Real coefficients.

We can simplify it with Complex coefficients by completing the square and using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \left(x + \frac{1}{2}\right)$ and $B = \frac{\sqrt{7}}{2} i$ as follows:

$0 = {x}^{2} + x + 2$

$\textcolor{w h i t e}{0} = {x}^{2} + 2 \left(x\right) \left(\frac{1}{2}\right) + {\left(\frac{1}{2}\right)}^{2} + \frac{7}{4}$

$\textcolor{w h i t e}{0} = {\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{7}}{2} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x + \frac{1}{2}\right) - \frac{\sqrt{7}}{2} i\right) \left(\left(x + \frac{1}{2}\right) + \frac{\sqrt{7}}{2} i\right)$

$\textcolor{w h i t e}{0} = \left(x + \frac{1}{2} - \frac{\sqrt{7}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{7}}{2} i\right)$

Hence:

$x = - \frac{1}{2} \pm \frac{\sqrt{7}}{2} i$