Complex Numbers And Quadratic Equations question 686
Question: If the roots of the equations $ px^{2}+2qx+r=0 $ and $ qx^{2}-2\sqrt{pr}x+q=0 $ be real, then
Options:
A) $ p=q $
B) $ q^{2}=pr $
C) $ p^{2}=qr $
D) $ r^{2}=pr $
Show Answer
Answer:
Correct Answer: B
Solution:
Consider both equations $ px^{2}+2qx+r=0 $ … (i) and $ qx^{2}-2\sqrt{pr}.x+q=0 $ … (ii) Since, both the equations are quadratic and have real roots, therefore from equation (1), we have
$ \therefore 4q^{2}-4pr\ge 0 $ (using discriminant)
$ \Rightarrow q^{2}\ge pr $ … (iii) and from second equation $ 4pr - 4q^{2} \ge 0 $
$ \Rightarrow pr\ge q^{2} $ … (iv) From eqs. (iii) and (iv) we get $ {q^{2}}= pr. $
BETA
