Answer
$\left\{-3,-2\sqrt{2},2\sqrt{2},3\right\}$
Work Step by Step
Using the properties of equality, the given expression, $
z^4+72=17z^2
,$ is equivalent to
\begin{align*}
z^4-17z^2+72&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z^2-9)(z^2-8)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z^2-9=0 & z^2-8=0
\\
z^2=9 & z^2=8
\\
z=\pm\sqrt{9} & z=\pm\sqrt{8}
\\
z=\pm3 & z=\pm\sqrt{4\cdot2}
\\
& z=\pm2\sqrt{2}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }z=\pm3: & \text{If }z=\pm2\sqrt{2}
\\\\
(\pm3)^4+72\overset{?}=17(\pm3)^2 &
(\pm2\sqrt{2})^4+72\overset{?}=17(\pm2\sqrt{2})2
\\
81+72\overset{?}=17(9) &
(16\cdot4)+72\overset{?}=17(4\cdot2)
\\
153\overset{\checkmark}=153 &
64+72\overset{?}=17(8)
\\
&
136\overset{\checkmark}=136
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
z^4+72=17z^2
$ is $\left\{-3,-2\sqrt{2},2\sqrt{2},3\right\}$.
