Question

If the sum of the roots of the quadratic equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ is zero. Show that the product of the roots is $-$$\frac{p^2+q^2}{2}$

Answer 1

$$\begin{gathered} \mathrm{First},\mathrm{we}\text{will}\mathrm{rewrite}\mathrm{the}\mathit{}\mathrm{given}\mathrm{quadratic}\mathrm{equation}\mathrm{in}\mathrm{its}\mathrm{standard}\mathrm{form}: \\ \frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r} \\ =>\frac{x+q+x+p}{(x+p)(x+q)}=\frac{1}{r} \\ =>(x+p)(x+q)=r(2x+p+q) \\ =>x^2+(p+q)x+pq=2rx+pr+qr \\ =>x^2+(p+q-2r)x+(pq-qr-rp)=0 \\ \mathrm{Let}\alpha \mathrm{and}\beta \text{be}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{this}\mathrm{quadratic}\mathrm{equation}. \\ \mathrm{Then}, \\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}=-\frac{p+q-2r}{1}=-\left(2r-p-q\right)\mathrm{and} \\ \mathrm{\alpha \beta }=\frac{c}{a}=\frac{pq-qr-rp}{1}=pq-r(q+p) \\ \mathrm{Since}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{is}\mathrm{zero},\alpha +\beta =0. \\ =>2r-p-q=0 \\ =>2r=p+q \\ =>r=\frac{p+q}{2} \\ \mathrm{On}\mathrm{substituting}\mathrm{this}\mathrm{value}\mathrm{of}r\mathrm{in}\mathrm{\alpha \beta }=pq-r(q+p),\mathrm{we}\mathrm{get}: \\ \mathrm{\alpha \beta }=pq-\left(\frac{p+q}{2}\right)(q+p) \\ =\frac{2pq-(p+q{)}^2}{2} \\ =\frac{2pq-p^2-q^2-2pq}{2} \\ =\frac{-p^2-q^2}{2} \end{gathered}$$