Answer
$\left\{-\dfrac{27}{8},-1,1,\dfrac{27}{8}\right\}$
Work Step by Step
Let $z=
x^{2/3}
$. Then the given equation, $
4x^{4/3}-13x^{2/3}+9=0
,$ is equivalent to
\begin{align*}
4\left(x^{2/3}\right)^2-13x^{2/3}+9&=0
\\
4z^2-13z+9&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-1)(4z-9)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-1=0 & 4z-9=0
\\
z=1 & 4z=9
\\\\
& z=\dfrac{9}{4}
.\end{array}
Since $z=
x^{2/3}
,$ by back substitution, then
\begin{array}{l|r}
x^{2/3}=1 & x^{2/3}=\dfrac{9}{4}
\\\\
\left(x^{\frac{2}{3}}\right)^3=(1)^3 & \left(x^{\frac{2}{3}}\right)^3=\left(\dfrac{9}{4}\right)^3
\\\\
x^2=1 & x^2=\dfrac{729}{64}
\\\\
x=\pm\sqrt{1} & x=\pm\sqrt{\dfrac{729}{64}}
\\\\
x=\pm1 & x=\pm\dfrac{27}{8}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\pm1: & \text{If }x=\pm\dfrac{27}{8}
\\\\
4(\pm1)^{4/3}-13(\pm1)^{2/3}+9\overset{?}=0 &
4\left(\pm\dfrac{27}{8}\right)^{4/3}-13\left(\pm\dfrac{27}{8}\right)^{2/3}+9\overset{?}=0
\\\\
4\left(\sqrt[3]{\pm1}\right)^4-13\left(\sqrt[3]{\pm1}\right)^2+9\overset{?}=0 &
4\left(\sqrt[3]{\pm\dfrac{27}{8}}\right)^4-13\left(\sqrt[3]{\pm\dfrac{27}{8}}\right)^2+9\overset{?}=0
\\\\
4\left(\pm1\right)^4-13\left(\pm1\right)^2+9\overset{?}=0 &
4\left(\pm\dfrac{3}{2}\right)^4-13\left(\pm\dfrac{3}{2}\right)^2+9\overset{?}=0
\\\\
4\left(1\right)-13\left(1\right)+9\overset{?}=0 &
4\left(\dfrac{81}{16}\right)-13\left(\dfrac{9}{4}\right)+9\overset{?}=0
\\\\
4-13+9\overset{?}=0 &
\dfrac{81}{4}-\dfrac{117}{4}+9\overset{?}=0
\\\\
0\overset{\checkmark}=0 &
-\dfrac{36}{4}+9\overset{?}=0
\\\\
&
-9+9\overset{?}=0
\\
&
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
4x^{4/3}-13x^{2/3}+9=0
$ is $\left\{-\dfrac{27}{8},-1,1,\dfrac{27}{8}\right\}$.
