$\frac{1}{x}+\frac{1}{p-x}=\frac{1}{q}$ সমীকরণের মূলদ্বয়ের অন্তর r হলে p কে q ও r এর মাধ্যমে প্রকাশ কর ।

Solve:$\frac{1}{x}+\frac{1}{p-x}=\frac{1}{q}$

$\begin{array}{l} \Rightarrow \frac{p-x+x}{x(p-x)}=\frac{1}{q} \Rightarrow p q=p x-x^{2} \\ \Rightarrow x^{2}-p x+p q=0 \end{array}$

এখন মনে করি, $\mathrm{x}^{2}-\mathrm{px}+\mathrm{pq}=0$ সমীকরণের মূলদ্বয় $\alpha$ এবং $\alpha+r$.

$\begin{array}{l} \therefore \alpha+\alpha+r=p \Rightarrow 2 \alpha=p-r \\ \Rightarrow \alpha=\frac{1}{2}(p-r) \cdots \cdots \text { (i) and } \\ \alpha(\alpha+r)=p q \Rightarrow \alpha^{2}+r \alpha=p q \\ \Rightarrow \frac{1}{4}(p-r)^{2}+\frac{1}{2} r(p-r)-p q=0 \\ \Rightarrow p^{2}-2 p r+r^{2}+2 p r-2 r^{2}-4 p q=0 \\ \Rightarrow p^{2}-4 p q-r^{2}=0 \\ \therefore p=\frac{4 q \pm \sqrt{16 q^{2}-4 \cdot 1 \cdot\left(-r^{2}\right)}}{2.1} \\ =\frac{4 q \pm 2 \sqrt{4 q^{2}+r^{2}}}{2}=2 q \pm \sqrt{4 q^{2}+r^{2}} \\ \end{array}$