Question

determine the solution set of the quadratic inequality: $4x^{2}-25leq0$. your answer should be in interval notation.
$(-\infty,-\frac{5}{2})cup(\frac{5}{2},\infty)$
$(-\frac{5}{2},\frac{5}{2})$
$(-\infty,-\frac{5}{2}cup\frac{5}{2},\infty)$
$-\frac{5}{2},\frac{5}{2}$

Explanation:

Step1: Factor the quadratic

We have $4x^{2}-25=(2x - 5)(2x+5)\leq0$.

Step2: Find the roots

Set each factor equal to zero: $2x - 5=0$ gives $x=\frac{5}{2}$, and $2x + 5=0$ gives $x=-\frac{5}{2}$.

Step3: Test intervals

We have three intervals to test: $(-\infty,-\frac{5}{2})$, $(-\frac{5}{2},\frac{5}{2})$, and $(\frac{5}{2},\infty)$.

• For $x<-\frac{5}{2}$, let $x=-3$. Then $(2(-3)-5)(2(-3)+5)=( - 6 - 5)(-6 + 5)=(-11)\times(-1)=11>0$.
• For $-\frac{5}{2}
• For $x>\frac{5}{2}$, let $x = 3$. Then $(2\times3-5)(2\times3+5)=(6 - 5)(6 + 5)=1\times11=11>0$.

Since we want $(2x - 5)(2x+5)\leq0$, the solution is the interval that makes the expression non - positive, which is $[-\frac{5}{2},\frac{5}{2}]$.

Answer:

$[-\frac{5}{2},\frac{5}{2}]$