Let $\mathrm{a}, \mathrm{b}$ be the roots of $x^{2}+p / x^{2}+q=0$
$\mathrm{So}, \mathrm{a}+\mathrm{b}=-\mathrm{p} / \mathrm{x}^{2}, \mathrm{ab}=\mathrm{q}$...(i)
Let $c, d$ be the roots of $x^{2}+b x+m=0$
So, $\mathrm{c}+\mathrm{d}=-1, \mathrm{~cd}=\mathrm{m}$...(ii)
Given that roots of both the equations are in the
Same ratio
So, $\frac{a}{b}=\frac{c}{d} \quad$...(iii)
$\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{c}}\ldots(\mathrm{iv})$
$(\mathrm{iii})+(\mathrm{iv}) \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{c}}$
$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{ab}}=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}}{\mathrm{ed}} \Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{ab}}+2=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}}{\mathrm{~cd}}+2$
$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{\mathrm{ab}}=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}+2 \mathrm{~cd}}{\mathrm{~cd}}$
$\Rightarrow \frac{(\mathrm{a}+\mathrm{b})^{2}}{\mathrm{ab}}=\frac{(\mathrm{c}+\mathrm{d})^{2}}{\mathrm{ed}}$
$\Rightarrow \frac{(-\mathrm{P})^{2}}{\mathrm{q}}=\frac{(-1)^{2}}{\mathrm{~m}} \quad($ from (i) and(ii))
$\Rightarrow \mathrm{P}^{2} \mathrm{~m}=1^{2} \mathrm{q}$