Answer
$\left( \frac{3}{2}+\frac{\sqrt{33}}{2},0 \right),\left( \frac{3}{2}-\frac{\sqrt{33}}{2},0 \right),\left( 4,0 \right)\text{ and }\left( -1,0 \right)$
Work Step by Step
$f\left( x \right)={{\left( {{x}^{2}}-3x \right)}^{2}}-10\left( {{x}^{2}}-3x \right)+24$.
The x-intercept occurs when $f\left( x \right)=0$,
${{\left( {{x}^{2}}-3x \right)}^{2}}-10\left( {{x}^{2}}-3x \right)+24=0$ …… (1)
Let $u=\left( {{x}^{2}}-3x \right)$ and ${{u}^{2}}={{\left( {{x}^{2}}-3x \right)}^{2}}$.
Substitute the values of $u$ and ${{u}^{2}}$ in equation (1),
${{u}^{2}}-10u+24=0$
Now, factor the equation.
$\left( u-6 \right)\left( u-4 \right)=0$
Apply the zero product rule,
$\left( u-6 \right)=0$or$\left( u-4 \right)=0$
If $\left( u-6 \right)=0$
$\begin{align}
& u-6=0 \\
& u=6
\end{align}$
If $\left( u-4 \right)=0$
$\begin{align}
& u-4=0 \\
& u=4 \\
\end{align}$
Now, replace $u$ with $\left( {{x}^{2}}-3x \right)$
Solve for $\left( {{x}^{2}}-3x \right)=6$,
$\begin{align}
& \left( {{x}^{2}}-3x \right)=6 \\
& {{x}^{2}}-3x-6=0 \\
\end{align}$
$\begin{align}
& x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -6 \right)}}{2\left( 1 \right)} \\
& =\frac{3\pm \sqrt{9+24}}{2} \\
& =\frac{3\pm \sqrt{33}}{2}
\end{align}$
The roots are $\frac{3+\sqrt{33}}{2},\frac{3-\sqrt{33}}{2}$. …… (2)
Solve for $\left( {{x}^{2}}-3x \right)=4$,
$\begin{align}
& \left( {{x}^{2}}-3x \right)=4 \\
& {{x}^{2}}-3x-4=0 \\
\end{align}$
$\begin{align}
& x=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -4 \right)}}{2\left( 1 \right)} \\
& =\frac{3\pm \sqrt{9+16}}{2} \\
& =\frac{3\pm \sqrt{25}}{2} \\
& =\frac{3\pm 5}{2}
\end{align}$
The roots are $4,-1$. …… (3)
Therefore from equations (2) and (3), the roots are $\frac{3+\sqrt{33}}{2},\frac{3-\sqrt{33}}{2},4\text{ and }-1$
Thus, the x-intercepts are $\left( \frac{3}{2}+\frac{\sqrt{33}}{2},0 \right),\left( \frac{3}{2}-\frac{\sqrt{33}}{2},0 \right),\left( 4,0 \right)\text{ and }\left( -1,0 \right)$.
