Question

Show that the roots of the equation $x^2+px-q^2=0$ are real for all real value of p and q.

Answer 1

$$\begin{gathered} \text{Given:} \\ {x}^2+px-q^2=0 \\ \text{Here,} \\ a=1,b=p\mathrm{and}c=-q^2 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ =p^2-4\times 1\times (-q^2) \\ =(p^2+4q^2)>0 \\ D>0\text{for all real values of}p\text{and}q. \\ \text{Thus, the roots of the equation are real}\text{.} \end{gathered}$$