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

Aug 6, 2016

#### Answer:

$x = - 1 - \sqrt{6}$ or $x = - 1 + \sqrt{6}$

#### Explanation:

${x}^{2} + 2 x - 5 = 0$

Now, recalling the identity ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ and comparing it with ${x}^{2} + 2 x$, we need to add and subtract ${\left(\frac{2}{1}\right)}^{2}$ to complete square. Hence ${x}^{2} + 2 x - 5 = 0$ is

${x}^{2} + 2 x + 1 - 1 - 5 = 0$ or

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

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

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

i.e. $x = - 1 - \sqrt{6}$ or $x = - 1 + \sqrt{6}$