Question

solve the quadratic inequality (4x^{2}+20x + 9leq0) by factoring. (1 point)
(-\frac{9}{2},-\frac{1}{2}
-\frac{9}{4},-1
(-infty,-9cup-\frac{1}{4},infty)
(-infty,-\frac{1}{2}cup\frac{9}{2},infty))

Explanation:

Step1: Factor the quadratic expression

We factor $4x^{2}+20x + 9$ as $(2x + 1)(2x+9)$. So the inequality becomes $(2x + 1)(2x + 9)\leq0$.

Step2: Find the roots

Set each factor equal to zero:
$2x+1 = 0$ gives $x=-\frac{1}{2}$; $2x + 9=0$ gives $x=-\frac{9}{2}$.

Step3: Test intervals

We have three intervals to test: $(-\infty,-\frac{9}{2})$, $(-\frac{9}{2},-\frac{1}{2})$, and $(-\frac{1}{2},\infty)$.
For $x<-\frac{9}{2}$, let $x=-5$, then $(2(-5)+1)(2(-5)+9)=( - 9)(-1)=9>0$.
For $-\frac{9}{2}For $x>-\frac{1}{2}$, let $x = 0$, then $(2(0)+1)(2(0)+9)=(1)(9)=9>0$.
The solution of the inequality is the interval where the expression is non - positive, which is $[-\frac{9}{2},-\frac{1}{2}]$.

Answer:

A. $[-\frac{9}{2},-\frac{1}{2}]$