# How do you solve 7x^2 - 12x + 16 = 0 ?

Jan 25, 2017

$x = \frac{6}{7} \pm \frac{2}{7} \sqrt{19} i$

#### Explanation:

$7 {x}^{2} - 12 x + 16 = 0$

is in the form:

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

with $a = 7$, $b = - 12$ and $c = 16$

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

$\Delta = {b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \left(7\right) \left(16\right) = 144 - 448 = - 304$

Since $\Delta < 0$, this quadratic equation has no Real roots, only Complex one.

We can still find the roots by completing the square.

The difference of squares identity can be written:

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

We use this with $A = \left(7 x - 6\right)$ and $B = 2 \sqrt{19} i$ as follows:

$0 = 7 \left(7 {x}^{2} - 12 x + 16\right)$

$\textcolor{w h i t e}{0} = 49 {x}^{2} - 84 x + 112$

$\textcolor{w h i t e}{0} = 49 {x}^{2} - 84 x + 36 + 76$

$\textcolor{w h i t e}{0} = {\left(7 x\right)}^{2} - 2 \left(7 x\right) \left(6\right) + {\left(6\right)}^{2} + {\left(2 \sqrt{19}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(7 x - 6\right)}^{2} - {\left(2 \sqrt{19} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(7 x - 6\right) - 2 \sqrt{19} i\right) \left(\left(7 x - 6\right) + 2 \sqrt{19} i\right)$

$\textcolor{w h i t e}{0} = \left(7 x - 6 - 2 \sqrt{19} i\right) \left(7 x - 6 + 2 \sqrt{19} i\right)$

Hence:

$x = \frac{1}{7} \left(6 \pm 2 \sqrt{19} i\right) = \frac{6}{7} \pm \frac{2}{7} \sqrt{19} i$