How do you solve the quadratic equation by completing the square: x^2 + 4x = 21?

Aug 12, 2015

${x}_{1 , 2} = - 2 \pm 5$

Explanation:

To solve this quadratic by completing the square, you need to use the coefficient of the $x$-term to help you find a number that when added to both sides of the equation will allow you to write the left side as the square of a binomial.

More specifically, you need to divide the coefficient of the $x$-term by $2$, the nsquare the result

${\left(\frac{4}{2}\right)}^{2} = {2}^{2} = 4$

Add this term to both sides of the equation to get

${x}^{2} + 4 x + 4 = 21 + 4$

Now, the left side of the equaation can be written as

${x}^{2} + 4 x + 4 = {x}^{2} + 2 \cdot \left(2\right) \cdot x + {\left(2\right)}^{2} = {\left(x + 2\right)}^{2}$

This means that you now have

${\left(x + 2\right)}^{2} = 25$

Take the square root of both sides

$\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{25}$

$x + 2 = \pm 5$

$x = - 2 \pm 5 = \left\{\begin{matrix}{x}_{1} = - 2 - 5 = - 7 \\ {x}_{2} = - 2 + 5 = 3\end{matrix}\right.$