# How do you solve x^2+x-72=0?

Dec 23, 2015

Find a pair of factors of $72$ which differ by $1$, hence:

${x}^{2} + x - 72 = \left(x + 9\right) \left(x - 8\right)$

which has zeros $x = 8$ and $x = - 9$

#### Explanation:

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

In our case we want $a b = 72$ and $a - b = 1$

It should not take long to find that $a = 9$ and $b = 8$ works.

Dec 23, 2015

Complete the square to find zeros $x = 8$ and $x = - 9$

#### Explanation:

Alternatively, complete the square as follows:

Note that ${\left(x + \frac{1}{2}\right)}^{2} = {x}^{2} + x + \frac{1}{4}$

We also use the difference of squares identity:

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

with $a = x + \frac{1}{2}$ and $b = \frac{17}{2}$

So:

${x}^{2} + x - 72$

$= {x}^{2} + x + \frac{1}{4} - \frac{1}{4} - 72 = {\left(x + \frac{1}{2}\right)}^{2} - \frac{289}{4}$

$= {\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{17}{2}\right)}^{2}$

$= \left(\left(x + \frac{1}{2}\right) - \frac{17}{2}\right) \left(\left(x + \frac{1}{2}\right) + \frac{17}{2}\right)$

$= \left(x - 8\right) \left(x + 9\right)$

which has zeros $x = 8$ and $x = - 9$