# How do solve x^2+21>10x and write the answer as a inequality and interval notation?

Nov 6, 2016

Rewrite as a quadratic equation, solve, and use test points to see in what interval the solution lies.

${x}^{2} + 21 = 10 x$

${x}^{2} - 10 x + 21 = 0$

$\left(x - 7\right) \left(x - 3\right) = 0$

$x = 7 \mathmr{and} 3$

TEST POINT 1: $x = 2$

2^2 + 21 >^? 10(2)

25 > 20" "color(green)(√)

TEST POINT 2: $x = 4$

4^2 + 21 >^? 10(4)

$37 {>}^{\emptyset} 40 \text{ } \textcolor{red}{\times}$

TEST POINT 3: $x = 8$

8^2 + 21>^?10(8)

85 > 80" "color(green)(√)

So, the solution is $x < 3 \mathmr{and} x > 7$.

In interval notation, this would be $\left(- \infty , 3\right)$ and $\left(7 , \infty\right)$.

Hopefully this helps!