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

Jun 30, 2015

Force a perfect square trinomial on the left side. Take the square root of both sides. Solve for $x$, which will have two values.

#### Explanation:

Completing the square involves forcing a perfect square trinomial on the left side of the equation, then solving for $x$.

The form for a perfect square is ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$.

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

Add $5$ to both sides of the equation.

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

Divide the coefficient of the $x$ value by $2$, then square the result.

2/2=1; ${1}^{2} = 1$

Add the result to both sides.

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

${x}^{2} + 2 x + 1 = 6$

The left side is now a perfect square trinomial.

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

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

Take the square root on both sides.

$x + 1 = \pm \sqrt{6}$

Subtract $1$ from both sides.

$x = \sqrt{6} - 1$

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