Question

$p^2{x}^2+(p^2-q^2)x-q^2=0$

Answer 1

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