We have,
$$
x^2-3 a x+14=0 \text { and } x^2+2 a x-16=0
$$

Let the common roots is $\alpha$.
Then, $\alpha^2-3 a \alpha+14=0$ and $\alpha^2+2 a \alpha-16=0$
by Cramer rule,
$$
\frac{\alpha^2}{48 a-28 a}=\frac{\alpha}{14+16}=\frac{1}{2 a+3 a}
$$


$\Rightarrow \quad \frac{\alpha^2}{20 a}=\frac{\alpha}{30}=\frac{1}{5 a}$
So, $\quad \frac{\alpha^2}{20 a}=\frac{\alpha}{30}$ and $\frac{\alpha}{30}=\frac{1}{5 a} \Rightarrow \alpha=\frac{20 a}{30}$
and $\quad \alpha=\frac{30}{5 a} \quad \alpha=\frac{2}{3} a, \alpha=\frac{6}{a}$
So, $\quad \frac{2 a}{3}=\frac{6}{a} \quad \Rightarrow 2 a^2=18$
$\Rightarrow \quad a^2=9 \quad \Rightarrow a= \pm 3$
Now, $a^4+a^2$
$$
=( \pm 3)^4+( \pm 3)^2=81+9=90
$$