Answer
$\left\{-\dfrac{4}{3},-1,1,\dfrac{4}{3}\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
9x^4-25x^2+16=0
,$ is equivalent to
\begin{align*}
(9x^2-16)(x^2-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
9x^2-16=0 & x^2-1=0
\\
9x^2=16 & x^2=1
\\\\
x^2=\dfrac{16}{9} & x=\pm\sqrt{1}
\\\\
x=\pm\sqrt{\dfrac{16}{9}} & x=\pm1
\\\\
x=\pm\dfrac{4}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\pm\dfrac{4}{3}: & \text{If }x=\pm1
\\\\
9\left(\pm\dfrac{4}{3}\right)^4-25\left(\pm\dfrac{4}{3}\right)^2+16\overset{?}=0 &
9(\pm1)^4-25(\pm1)^2+16\overset{?}=0
\\\\
9\left(\dfrac{256}{81}\right)-25\left(\dfrac{16}{9}\right)+16\overset{?}=0 &
9(1)-25(1)+16\overset{?}=0
\\\\
\dfrac{256}{9}-\dfrac{400}{9}+16\overset{?}=0 &
9-25+16\overset{?}=0
\\\\
\dfrac{256}{9}-\dfrac{400}{9}+\dfrac{144}{9}\overset{?}=0 &
-16+16\overset{?}=0
\\\\
-\dfrac{144}{9}+\dfrac{144}{9}\overset{?}=0 &
0\overset{\checkmark}=0
\\\\
0\overset{?}=0 &
.\end{array}
Since both solutions satisfy the original equation, then the solution set of the equation $
9x^4-25x^2+16=0
$ is $\left\{-\dfrac{4}{3},-1,1,\dfrac{4}{3}\right\}$.
