Question
If the roots of the equations $$px^2 + 2qx + r = 0$$ and $$q{x^2} - 2\sqrt {pr} x + q = 0$$ be real, then
A.
$$p = q$$
B.
$$q^2 = pr$$
C.
$$p^2 = qr$$
D.
$$r^2 = pr$$
Answer :
$$q^2 = pr$$
Solution :
Consider both equations
$$\eqalign{
& p{x^2} + 2qx + r = 0\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{and, }}q{x^2} - 2\sqrt {pr} .x + q = 0\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Since, both the equations are quadratic and have real roots, therefore from equation (i), we have
$$\eqalign{
& \therefore 4{q^2} - 4pr \geqslant 0\left( {{\text{using discriminant}}} \right) \cr
& \Rightarrow {q^2} \geqslant pr\,\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
and from second equation $$4pr - 4{q^2} \geqslant 0$$
$$ \Rightarrow pr \geqslant {q^2}\,\,\,.....\left( {{\text{iv}}} \right)$$
From equations (iii) and (iv) we get, $$q^2 = pr.$$