Since, $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+p x+q=0$
$\therefore \quad$ Sum of roots $=-p$
$\Rightarrow \quad(\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-P$
$\Rightarrow$
$\mathbf{2}\alpha=-p \Rightarrow \alpha=-\frac{\underline{p}}{2}$
and produet of roots $=\mathrm{q}$
$(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta})=q$
$\Rightarrow \alpha^{2}-\beta=q$
$\Rightarrow \quad \beta=\alpha^{2}-q=\left(-\frac{p}{2}\right)^{2}-q=\frac{p^{2}}{4}-q$
$\Rightarrow \quad p^{2}-4 q=4 \beta$
$\therefore$ The given equation $\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0$
$4 \beta\left(p^{2} x^{2}+4 p x\right)-16\left(\alpha^{2}-\beta\right)=0$
$\left[\alpha^{2}-\beta=q\right]$
$\beta\left(4 a^{2} x^{2}-8 x x\right)-4\left(a^{2}-\beta\right)=0$
$\alpha^{2} \beta x^{2}-2 x f\left(x+\beta=\alpha^{2}\right.$
$\Rightarrow \quad(\alpha x \sqrt{\beta}-\sqrt{\beta})^{2}=\alpha^{2}$
$\Rightarrow \quad \operatorname{cox} \sqrt{\beta}-\sqrt{\beta}=\pm \alpha$
$\therefore$
$x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$
$\Rightarrow x=\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$