Solve the problem.
-If $h ( x ) = x ^ { 2 } + 2 x - 8$, solve $h ( x ) 0$.
A) $( - \infty , - 4 )$
B) $( - 4,2 )$
C) $( - \infty , - 4 )$ or $( 2 , \infty )$
D) $( 2 , \infty )$

Question

Solve the problem.
-If $h ( x ) = x ^ { 2 } + 2 x - 8$, solve $h ( x ) > 0$.
A) $( - \infty , - 4 )$
B) $( - 4,2 )$
C) $( - \infty , - 4 )$ or $( 2 , \infty )$
D) $( 2 , \infty )$

Quizplus · Accepted Answer

To solve $h(x) > 0$, we first find the roots of the quadratic equation $x^2 + 2x - 8 = 0$. Factoring or using the quadratic formula, we find the roots to be $x = -4$ and $x = 2$. The quadratic opens upwards (since the coefficient of $x^2$ is positive), so $h(x) > 0$ outside the interval between the roots, which gives us $(-\infty, -4)$ and $(2, \infty)$.

Solve the Problem h(x)=x2+2x−8h ( x ) = x ^ { 2 } + 2 x - 8h(x)=x2+2x−8

Solve the Problem $h ( x ) = x ^ { 2 } + 2 x - 8$