# How do you solve x^2+4x= -1?

May 20, 2016

$x = - 2 \setminus \pm \setminus \sqrt{3}$

#### Explanation:

There are several ways to do it, by I would prefer in this case to use the completing the square method to find the root of the quadratic ${x}^{2} + 4 x = - 1$.

• In completing the square, we need to first make sure whether the a value of the quadratic equation equals to 1. If not, then you need to divide every coefficient and constant by the $a$ value to make sure that the $a$ value is 1. In this case, this is not required.

• Now we need to divide the $b$ term by 2 and then square it. Do this to both sides.
${x}^{2} + 4 x + {\left(\setminus \frac{4}{2}\right)}^{2} = - 1 + {\left(\setminus \frac{4}{2}\right)}^{2}$
${x}^{2} + 4 x + 4 = - 1 + 4$
${x}^{2} + 4 x + 4 = 3$

• Put ${x}^{2} + 4 x + 4$ into a binomial.
${\left(x + 2\right)}^{2} = 3$

• To solve for $x$, find the square root of both sides.
$\setminus \sqrt{{\left(x + 2\right)}^{2}} = \setminus \sqrt{3}$
$x + 2 = \setminus \pm \setminus \sqrt{3}$
Notice the $\setminus \pm$ in front of the square root of 3. This is because the square root of 3 can be both a positive and a negative (a negative times a negative is a positive).

• Isolate $x$ by subtracting 2 from both sides.
$x = - 2 \setminus \pm \setminus \sqrt{3}$

Answer: $x = - 2 \setminus \pm \setminus \sqrt{3}$