Answer
${{x}^{2}}+9=0$.
Work Step by Step
$3i$ and $-3i$.
That is, $x=3i$ and $x=-3i$.
Now, get 0’s on one side for both the solutions.
For $x=3i$, subtract $3i$ on both sides:
That is,
$\begin{align}
& x=3i \\
& x-3i=3i-3i \\
& x-3i=0
\end{align}$
Now, for $x=-3i$ add $3i$ on both sides:
That is,
$\begin{align}
& x=3-i \\
& x+3i=3i-3i \\
& x+3i=0
\end{align}$
Multiply both the products $x-3i$ and $x+3i$:
That is,
$\begin{align}
& \left( x-3i \right)\left( x+3i \right)=0 \\
& {{x}^{2}}-{{\left( 3i \right)}^{2}}=0
\end{align}$
Consider the term, ${{\left( 3i \right)}^{2}}$.
The term ${{\left( 3i \right)}^{2}}$ is equal to $-9$ as ${{i}^{2}}=-1$.
Therefore,
$\begin{align}
& {{x}^{2}}-{{\left( 3i \right)}^{2}}=0 \\
& {{x}^{2}}-\left( -9 \right)=0 \\
& {{x}^{2}}+9=0
\end{align}$
Therefore, the quadratic equation is ${{x}^{2}}+9=0$.
