Answer
True.
The values of x in the equation $3{{x}^{2}}+48=0$ are $\pm 4i$.
Work Step by Step
Consider the equation.
$3{{x}^{2}}+48=0$ …… (1)
Subtract $48$ on both sides.
$\begin{align}
& 3{{x}^{2}}+48-48=0-48 \\
& 3{{x}^{2}}=-48 \\
& {{x}^{2}}=-\frac{48}{3} \\
& {{x}^{2}}=-16
\end{align}$
Apply the square root property.
$\begin{align}
& x=\sqrt{-16} \\
& =\pm \sqrt{-1}\cdot \sqrt{16}
\end{align}$
Substitute $\sqrt{-1}=i$.
$\begin{align}
& x=i\sqrt{16} \\
& =\pm 4i
\end{align}$
Check,
Substitute $\pm 4i$ in equation (1).
$\begin{align}
3{{\left( \pm 4i \right)}^{2}}+48\overset{?}{\mathop{=}}\,0 & \\
3{{\left( \pm 4 \right)}^{2}}{{\left( i \right)}^{2}}+48\overset{?}{\mathop{=}}\,0 & \\
3\left( 16 \right)\left( -1 \right)+48\overset{?}{\mathop{=}}\,0 & \\
-48+48\overset{?}{\mathop{=}}\,0 & \\
\end{align}$
Solve further,
$\begin{align}
-48+48\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
It is true.
