Answer
$\left\{-2\sqrt{3},-2,2,2\sqrt{3},\right\}$
Work Step by Step
Using the properties of equality, the given expression, $
x^4+48=16x^2
,$ is equivalent to
\begin{align*}
x^4-16x^2+48&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x^2-4)(x^2-12)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x^2-4=0 & x^2-12=0
\\
x^2=4 & x^2=12
\\
x=\pm\sqrt{4} & x=\pm\sqrt{12}
\\
x=\pm2 & x=\pm\sqrt{4\cdot3}
\\
& x=\pm2\sqrt{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\pm2: & \text{If }x=\pm2\sqrt{3}
\\\\
(\pm2)^4+48\overset{?}=16(\pm2)^2 &
(\pm2\sqrt{3})^4+48\overset{?}=16(\pm2\sqrt{3})^2
\\
16+48\overset{?}=16(4) &
16(9)+48\overset{?}=16(4\cdot3)
\\
64\overset{\checkmark}=64 &
144+48\overset{?}=16(12)
\\
&
192\overset{\checkmark}=192
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
9x^4-25x^2+16=0
$ is $\left\{-2\sqrt{3},-2,2,2\sqrt{3},\right\}$.
