# How do you solve the inequality 3 <= x^2 - 8x + 15?

May 13, 2017

Because the coefficient of the ${x}^{2}$ term is greater than 0, we know that the domain between the roots will cause the quadratic to be less than 0. We shall include all values of x except this region.

#### Explanation:

Given: $3 \le {x}^{2} - 8 x + 15$

Subtract 3 from both sides:

$0 \le {x}^{2} - 8 x + 12$

Flip the inequality:

${x}^{2} - 8 x + 12 \ge 0$

Because the coefficient of the ${x}^{2}$ term is greater than 0, we know that the quadratic represents a parabola that opens upward. Therefore, the quadratic will be less than 0 between the two roots and greater than or equal to 0 elsewhere.

Let's find the roots by factoring:

$\left(x - 2\right) \left(x - 6\right) = 0$

$x = 2 \mathmr{and} x = 6$

Therefore, we know that the two domains that will cause the quadratic to be greater than or equal to 0 are:

$x \le 2$ and $x \ge 6$

Here is a graph to prove it:

graph{x^2-8x+12 [-10, 10, -5, 5]}