# How do you solve the inequality: |x^2-9x+15|>1 ?

Apr 19, 2016

$x \in \left(- \infty , 2\right) \cup \left(\frac{9 - \sqrt{17}}{2} , \frac{9 + \sqrt{17}}{2}\right) \cup \left(7 , \infty\right)$

#### Explanation:

By the definition of the absolute value of a number, this is equivalent to stating that

${x}^{2} - 9 x + 15 > 1$
or
${x}^{2} - 9 x + 15 < - 1$

We will solve each inequality separately.

For the first inequality, subtracting $1$ from each side, we obtain

${x}^{2} - 9 x + 14 > 0$

$\implies \left(x - 2\right) \left(x - 7\right) > 0$

$\implies x - 2 > 0$ and $x - 7 > 0$
or
$x - 2 < 0$ and $x - 7 < 0$

$\implies x > 7$
or
$x < 2$

$\implies x \in \left(- \infty , 2\right) \cup \left(7 , \infty\right)$

For the second inequality, we can add $1$ to both sides and use the quadratic formula to factor the left hand side.

${x}^{2} - 9 x + 16 < 0$

$\implies \left(x - \frac{9 + \sqrt{17}}{2}\right) \left(x - \frac{9 - \sqrt{17}}{2}\right) < 0$

Because $\left(x - \frac{9 + \sqrt{17}}{2}\right) < \left(x - \frac{9 - \sqrt{17}}{2}\right)$ and the inequality holds only when one of factors is negative and the other is positive, we must have that

$\left(x - \frac{9 + \sqrt{17}}{2}\right) < 0$ and $\left(x - \frac{9 - \sqrt{17}}{2}\right) > 0$

$\implies \frac{9 - \sqrt{17}}{2} < x < \frac{9 + \sqrt{17}}{2}$

$\implies x \in \left(\frac{9 - \sqrt{17}}{2} , \frac{9 + \sqrt{17}}{2}\right)$

To get our final solution set, we can take the union of both solutions we found above to obtain

$x \in \left(- \infty , 2\right) \cup \left(\frac{9 - \sqrt{17}}{2} , \frac{9 + \sqrt{17}}{2}\right) \cup \left(7 , \infty\right)$

Note that this matches what we would expect from the graph, as $| {x}^{2} - 9 x + 15 | - 1$ is negative on only two small intervals:

graph{|x^2-9x+15|-1 [-5.205, 14.795, -3.24, 6.76]}