Answer
$\left\{-\dfrac{3}{2},-1,1,\dfrac{3}{2}\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
4q^4-13q^2+9=0
,$ is equivalent to
\begin{align*}
(4q^2-9)(q^2-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
4q^2-9=0 & q^2-1=0
\\
4q^2=9 & q^2=1
\\\\
q^2=\dfrac{9}{4} & q=\pm\sqrt{1}
\\\\
q=\pm\sqrt{\dfrac{9}{4}} & q=\pm1
\\\\
q=\pm\dfrac{3}{2}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }q=\pm\dfrac{3}{2}: & \text{If }q=\pm1
\\\\
4\left(\pm\dfrac{3}{2}\right)^4-13\left(\pm\dfrac{3}{2}\right)^2+9\overset{?}=0 &
4(\pm1)^4-13(\pm1)^2+9\overset{?}=0=0
\\\\
4\left(\dfrac{81}{16}\right)-13\left(\dfrac{9}{4}\right)+9\overset{?}=0 &
4(1)-13(1)+9\overset{?}=0=0
\\\\
\dfrac{81}{4}-\dfrac{117}{4}+9\overset{?}=0 &
4-13+9\overset{?}=0=0
\\\\
\dfrac{81}{4}-\dfrac{117}{4}+\dfrac{36}{4}\overset{?}=0 &
0\overset{\checkmark}=0=0
\\\\
\dfrac{0}{4}\overset{?}=0
\\\\
0\overset{\checkmark}=0
.\end{array}
Since both solutions satisfy the original equation, then the solution set of the equation $
4q^4-13q^2+9=0
$ is $\left\{-\dfrac{3}{2},-1,1,\dfrac{3}{2}\right\}$.
