Question

If the equations px² + 2qx + r = 0 and $q{x}^2-2\sqrt{pr}x+q=0$ have real roots, then q² =____________.

Answer 1

$$\begin{gathered} \mathrm{Given}p{x}^2+2qx+r=0 \\ \mathrm{and}q{x}^2-2\sqrt{pr}x+q=0\mathrm{have}\mathrm{real}\mathrm{roots} \\ \mathrm{for}p{x}^2+2qx+r=0 \\ \mathrm{Since}\mathrm{roots}\mathrm{are}\mathrm{real} \\ \Rightarrow \mathrm{Discriminant}D\ge 0 \\ \mathrm{i}.\mathrm{e}.{\mathrm{b}}^2-4ac\ge 0 \\ \mathrm{i}.\mathrm{e}.{\left(2q\right)}^2-4\left(p\right)\left(r\right)\ge 0 \\ \mathrm{i}.\mathrm{e}.4q^2-4pr\ge 0 \\ \mathrm{i}.\mathrm{e}.\mathit{}{q}^2\ge pr...\left(\mathrm{i}\right) \\ \mathrm{also}\mathrm{for}q{x}^2-2\sqrt{pr}\mathit{}x+q=0 \\ D\ge 0\mathrm{i}.\mathrm{e}.b^2-4ac\ge 0 \\ \mathrm{i}.\mathrm{e}.4\mathrm{pr}-4\left(q\right)\left(q\right)\ge 0 \\ \mathrm{i}.\mathrm{e}.\mathit{}pr\ge q^2 \\ \mathrm{i}.\mathrm{e}.q^2\le pr\mathit{}...\left(\mathrm{ii}\right) \\ \therefore \mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ pr\le q^2\le pr \\ \Rightarrow q^2=pr \end{gathered}$$