Answer
$\left\{1,4\right\}$
Work Step by Step
Squaring both sides of the given equation, $
z=\sqrt{5z-4}
,$ results to
\begin{align*}\require{cancel}
(z)^2&=\left(\sqrt{5z-4}\right)^2
\\
z^2&=5z-4
\\
z^2-5z+4&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-4)(z-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-4=0 & z-1=0
\\
z=4 & z=1
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }z=4: & \text{If }z=1:
\\\\
4\overset{?}=\sqrt{5(4)-4} &
1\overset{?}=\sqrt{5(1)-4}
\\
4\overset{?}=\sqrt{20-4} &
1\overset{?}=\sqrt{5-4}
\\
4\overset{?}=\sqrt{16} &
1\overset{?}=\sqrt{1}
\\
4\overset{\checkmark}=4 &
1\overset{\checkmark}=1
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
z=\sqrt{5z-4}
$ is $\left\{1,4\right\}$.
