# How do you solve 2x^2 - x - 5 = 0 by completing the square?

Dec 3, 2017

$x = \frac{1}{4} \pm \frac{\sqrt{41}}{4}$

#### Explanation:

Premultiply by $8$ (to avoid some fractions), complete the square and use the difference of squares identity:

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

with $A = \left(4 x - 1\right)$ and $B = \sqrt{41}$ as follows:

$0 = 8 \left(2 {x}^{2} - x - 5\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} - 8 x - 40$

$\textcolor{w h i t e}{0} = {\left(4 x\right)}^{2} - 2 \left(4 x\right) + 1 - 41$

$\textcolor{w h i t e}{0} = {\left(4 x - 1\right)}^{2} - {\left(\sqrt{41}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x - 1\right) - \sqrt{41}\right) \left(\left(4 x - 1\right) + \sqrt{41}\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 1 - \sqrt{41}\right) \left(4 x - 1 + \sqrt{41}\right)$

So:

$4 x = 1 \pm \sqrt{41}$

and:

$x = \frac{1}{4} \pm \frac{\sqrt{41}}{4}$

Note that multiplying by $8$ has a couple of effects:

• It makes the leading term into a perfect square ready for completing the square.

• It makes the coefficient of $x$ into a multiple of twice the square root of the leading coefficient, thus avoiding having to use fractions until the end.