# How do you solve x² - x - 20 = 0?

Jul 13, 2015

Find two numbers whose product is $20$ and whose difference is $1$. The pair $5$, $4$ works.

Hence ${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$.

So solutions are $x = 5$ or $x = - 4$.

#### Explanation:

$\left(x - a\right) \left(x + b\right) = {x}^{2} - \left(a - b\right) x - a b$

Matching this against ${x}^{2} - x - 20$ we see that if we can find $a$ and $b$ such that $a - b = 1$ and $a b = 20$, then we can factor the quadratic into two linear terms.

The pair $a = 5$, $b = 4$ works, giving us

${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

This will be $0$ when $\left(x - 5\right) = 0$ or $\left(x + 4\right) = 0$, which is when $x = 5$ or $x = - 4$.