# How do you solve x^2 – 12x + 32 <12?

Dec 1, 2017

Start with equality, find the region and check where it is valid

#### Explanation:

We have the inequality as
${x}^{2} - 12 x + 32 < 12$
${x}^{2} - 12 x + 20 < 0$

${x}^{2} - 12 x + 20 = 0$
$\left(x - 10\right) \left(x - 2\right) = 0$
$x = 2 , 10$

So we have 3 ranges.
$\left(- \setminus \infty , 2\right) , \left(2 , 10\right) , \left(10 , \setminus \infty\right)$

Let us take some number from each region and check where it is less than 0.

$x = 0$ then we have ${0}^{2} - 12 \setminus \times 0 + 20 = 20 > 0$
$x = 11$ then we have ${11}^{2} - 12 \setminus \times 11 + 20 = 9 > 0$
$x = 3$ then we have ${3}^{2} - 12 \setminus \times 3 + 20 = - 7 < 0$

Hence the region where the inequality holds is $\left(2 , 10\right)$