# How do you solve x^2 - 3x - 28 = 0 by completing the square?

Jun 19, 2016

x is either 7 or $-$4

#### Explanation:

We have, ${x}^{2} - 3 x - 28 = 0$

This can be written as ${x}^{2} - 3 x = 28$

${x}^{2} - \left(2\right) \left(\frac{3}{2}\right) x = 28$ (the trick is to take out 2 from the coefficient of x)

Now adding the square of 3/2 on both sides.

${x}^{2} - \left(2\right) \left(\frac{3}{2}\right) x + {\left(\frac{3}{2}\right)}^{2} = 28 + {\left(\frac{3}{2}\right)}^{2}$

Carefully notice that the LHS becomes a square

${\left(x - \frac{3}{2}\right)}^{2} = 28 + \frac{9}{4}$

[A side note: if ${\left(f \left(x\right)\right)}^{2} = g \left(x\right)$ then $f \left(x\right) = \pm \sqrt{g \left(x\right)}$ ]

So, $x - \frac{3}{2} = \pm \sqrt{28 + \frac{9}{4}}$

$x - \frac{3}{2} = \pm \sqrt{\frac{121}{4}}$

$x - \frac{3}{2} = \pm \frac{11}{2}$

$x = \frac{3}{2} \pm \frac{11}{2}$

So x is either 7 or $-$4