#### Answer

True statement

#### Work Step by Step

$a{{x}^{2}}+bx+c=0$
An example of equation reducible to a quadratic is
${{x}^{4}}-8{{x}^{2}}-9=0$
Let ${{x}^{2}}=u$,
Then ${{x}^{4}}={{u}^{2}}$.
Substitute the values of ${{x}^{2\text{ }}}\text{and }{{x}^{4}}$ into the equation ${{x}^{4}}-8{{x}^{2}}-9=0$ as follows:
$\begin{align}
& {{x}^{4}}-8{{x}^{2}}-9=0 \\
& {{u}^{2}}-8u-9=0
\end{align}$
Factor ${{u}^{2}}-8u-9=0$,
$\begin{align}
& {{u}^{2}}-8u-9=0 \\
& \left( u-9 \right)\left( u+1 \right)=0
\end{align}$
Thus,
$\text{ }u=9\text{ or }u=-1$
Now, this is incomplete because the solution is being carried out for x and not u.
Now replace $u$ with ${{x}^{2}}$.
$\begin{align}
& \text{ }u=9\text{ or }u=-1 \\
& {{x}^{2}}\text{= 9 or }{{x}^{2}}=-1 \\
& x=\pm 3\text{ square can }\!\!'\!\!\text{ t be negative,no solution}\text{. } \\
\end{align}$
Thus, the final solution is $x=\pm \text{ }3$ and not $\text{ }u=9\text{ or }u=-1$.