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

Jul 28, 2015

${x}_{1} = 1 + \sqrt{6}$, ${x}_{2} = 1 - \sqrt{6}$

Explanation:

Start by writing your quadratic in the form

${x}^{2} + \frac{b}{a} x = - \frac{c}{a}$

In your case, $a = 1$ to begin with, so you have

${x}^{2} - 2 x = 5$

In order to solve this quadratic by completing the square, you need to write the left side of this equation as a square of a binomial by adding a term to both sides of the equation.

You can determine what this term must be by dividing the coefficient of the $x$-term by 2, then squaring the result.

In your case, you have

$\frac{- 2}{2} = \left(- 1\right)$, then

${\left(- 1\right)}^{2} = 1$

This means that the quadratic becomes

${x}^{2} - 2 x + 1 = 5 + 1$

The left side of the equation can now be written as

${x}^{2} - 2 x + 1 = {x}^{2} + 2 \cdot \left(- 1\right) + {\left(- 1\right)}^{2} = {\left(x - 1\right)}^{2}$

This means that you have

${\left(x - 1\right)}^{2} = 6$

Take the square root of both sides to get

$\sqrt{{\left(x - 1\right)}^{2}} = \sqrt{6}$

$x - 1 = \pm \sqrt{6} \implies {x}_{1 , 2} = 1 \pm \sqrt{6}$

The two solutions to your quadratic will thus be

${x}_{1} = \textcolor{g r e e n}{1 + \sqrt{6}}$ and ${x}_{2} = \textcolor{g r e e n}{1 - \sqrt{6}}$