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Question

# ${p}^{2}{x}^{2}+\left({p}^{2}-{q}^{2}\right)x-{q}^{2}=0$

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Solution

## $\text{Given:}\phantom{\rule{0ex}{0ex}}\text{}{p}^{2}{x}^{2}+\left({p}^{2}-{q}^{2}\right)x-{{q}^{2}}^{}=0\phantom{\rule{0ex}{0ex}}\mathrm{On}\text{comparing it with}a{x}^{2}+bx+c=0,\text{we get:}\phantom{\rule{0ex}{0ex}}\text{}a={p}^{2},b=\left({p}^{2}-{q}^{2}\right)\mathrm{and}c=-{q}^{2}\text{}\phantom{\rule{0ex}{0ex}}\mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}D=\left({b}^{2}-4ac\right)\text{}\phantom{\rule{0ex}{0ex}}\text{=}{\left({p}^{2}-{q}^{2}\right)}^{2}\text{}-4×{p}^{2}×\left(-{q}^{2}\right)\text{}\phantom{\rule{0ex}{0ex}}\text{=}{p}^{4}-2{p}^{2}{q}^{2}+{q}^{4}+4{p}^{2}{q}^{2}\text{}\phantom{\rule{0ex}{0ex}}={p}^{4}+2{p}^{2}{q}^{2}+{q}^{4}\text{}\phantom{\rule{0ex}{0ex}}={\left({p}^{2}+{q}^{2}\right)}^{2}\text{> 0}\phantom{\rule{0ex}{0ex}}\text{Hence, the roots of the equation are real.}\phantom{\rule{0ex}{0ex}}\mathrm{R}\text{oots}\mathit{\text{α}}\text{and}\mathit{\text{β}}\text{are given by:}\phantom{\rule{0ex}{0ex}}\alpha \text{=}\frac{-b+\sqrt{D}}{2a}\text{}=\frac{-\left({p}^{2}-{q}^{2}\right)+\sqrt{{\left({p}^{2}+{q}^{2}\right)}^{2}}}{2\text{}{p}^{2}}\text{}=\frac{2{q}^{2}}{2{p}^{2}}=\frac{{q}^{2}}{{p}^{2}}\phantom{\rule{0ex}{0ex}}\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-\left({p}^{2}-{q}^{2}\right)-\sqrt{{\left({p}^{2}+{q}^{2}\right)}^{2}}}{2\text{}{p}^{2}}\text{}=\frac{-2{p}^{2}}{2{p}^{2}}=-1\phantom{\rule{0ex}{0ex}}\text{Hence, the roots of the equation are}\frac{{q}^{2}}{{p}^{2}}\text{and}-1.$

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