Question

determine the solution set of the quadratic inequality: -x^2 + 4x + 5 > 0. your answer should be in interval notation. (-∞,-1 ∪ 5,∞) -1,5 (-∞, -1) ∪ (5,∞) (-1,5)

Explanation:

Step1: Rewrite the inequality

Multiply both sides by - 1 and reverse the inequality sign. We get $x^{2}-4x - 5<0$.

Step2: Factor the quadratic expression

Factor $x^{2}-4x - 5$ as $(x - 5)(x+1)<0$.

Step3: Find the roots

Set $(x - 5)(x + 1)=0$. The roots are $x = 5$ and $x=-1$.

Step4: Determine the solution interval

We test intervals $(-\infty,-1)$, $(-1,5)$ and $(5,\infty)$. For $x<-1$, let $x=-2$, then $(-2 - 5)(-2 + 1)=(-7)\times(-1)=7>0$. For $-15$, let $x=6$, then $(6 - 5)(6 + 1)=7>0$. So the solution of $(x - 5)(x + 1)<0$ is $-1

Answer:

$(-1,5)$