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

Feb 25, 2016

$\left(1 - \sqrt{6}\right)$ and $\left(1 + \sqrt{6}\right)$

#### Explanation:

To solve x^2−2x−5=0 using the completing the square method, one picks up terms containing ${x}^{2}$ and $x$. Here these are x^2−2x. Here coefficient of ${x}^{2}$ is one and that of $x$ is $- 2$, hence adding half of the square of latter i.e. adding ${\left(- 1\right)}^{2}$ (i..e. $1$) should make x^2−2x, a complete square.

Hence for solving, let us add and subtract $1$, and doing this the equation becomes

x^2−2x+1−6=0

or ${\left(x - 1\right)}^{2} - {\left(\sqrt{6}\right)}^{2} = 0$

i.e. $\left(x - 1 + \sqrt{6}\right) \left(x - 1 - \sqrt{6}\right) = 0$

i.e. $x = \left(1 - \sqrt{6}\right)$ or $x = \left(1 + \sqrt{6}\right)$