# How do you find the roots of x^2+x=56?

May 14, 2016

$x = 7$ or $x = - 8$

#### Explanation:

Note that ${x}^{2} + x = x \left(x + 1\right)$ and $56 = 7 \cdot 8$

So $x = 7$ is a root.

This is a quadratic, so it has another root.

$56 = \left(- 8\right) \cdot \left(- 7\right)$

So $x = - 8$ is the other root.

$\textcolor{w h i t e}{}$
Alternative method

Alternatively, we can subtract $56$ from both sides to get:

${x}^{2} + x - 56 = 0$

This is then in the form $a {x}^{2} + b x + c = 0$ with $a = 1$, $b = 1$ and $c = - 56$.

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \cdot 1 \cdot \left(- 56\right)\right)}}{2 \cdot 1}$

$= \frac{- 1 \pm \sqrt{1 + 224}}{2}$

$= \frac{- 1 \pm \sqrt{225}}{2}$

$= \frac{- 1 \pm 15}{2}$

We find:

$\frac{- 1 + 15}{2} = \frac{14}{2} = 7$

$\frac{- 1 - 15}{2} = \frac{- 16}{2} = - 8$