# How do you solve and write the following in interval notation: x^2-1x-30>0?

Apr 14, 2017

Solution : $x < - 5 \mathmr{and} x > 6$. In interval notation: $\left(- \infty , - 5\right) \cup \left(6 , \infty\right)$.

#### Explanation:

${x}^{2} - x - 30 > 0 \mathmr{and} {x}^{2} - 6 x + 5 x - 30 > 0 \mathmr{and} x \left(x - 6\right) + 5 \left(x - 6\right) > 0 \mathmr{and} \left(x + 5\right) \left(x - 6\right) > 0$
Critical points are $x = - 5 , x = 6$

When  x < -5; (x+5)(x-6) is (-)*(-) = (+) or >0

When  -5< x < 6; (x+5)(x-6) is (+)*(-) = (-) or <0

When  x > 6; (x+5)(x-6) is (+)*(+) = (+) or >0

Solution : $x < - 5 \mathmr{and} x > 6$. In interval notation: $\left(- \infty , - 5\right) \cup \left(6 , \infty\right)$. The graph also confirms the findings. graph{x^2-x-30 [-80, 80, -40, 40]}[Ans]