Solve the following

Solution:

Given $p x^{2}+2 q x+r=0$

and $q x^{2}-2 \sqrt{p r} x+q=0$ have real roots

for $p x^{2}+2 q x+r=0$

Since roots are real

$\Rightarrow$ Discriminant $D \geq 0$

i.e. $\mathrm{b}^{2}-4 a c \geq 0$

i. e. $(2 q)^{2}-4(p)(r) \geq 0$

i. e. $4 q^{2}-4 p r \geq 0$

i. e. $q^{2} \geq p r$    ....(i)

also for $q x^{2}-2 \sqrt{p r} x+q=0$

$D \geq 0$ i. e. $b^{2}-4 a c \geq 0$

i. e. $4 \mathrm{pr}-4(q)(q) \geq 0$

i. e. $p r \geq q^{2}$

i. e. $a^{2}

$\therefore$ from (i) and (ii)

$p r \leq q^{2} \leq p r$

$\Rightarrow q^{2}=p r$