Complex Numbers And Quadratic Equations question 761
Question: If the roots of the equation $ (p^{2}+q^{2})x^{2} $ $ -2q(p+r)x $ + $ (q^{2}+r^{2})=0 $ be real and equal, then $ p,q,r $ will be in
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ (p^{2}+q^{2})x^{2}-2q(p+r)x+(q^{2}+r^{2})=0 $ Roots are real and equal, then $ 4q^{2}{{(p+r)}^{2}}-4(p^{2}+q^{2})(q^{2}+r^{2})=0 $
Þ $ q^{2}(p^{2}+r^{2}+2pr)-(p^{2}q^{2}+p^{2}r^{2}+q^{4}+q^{2}r^{2})=0 $
Þ $ q^{2}p^{2}+q^{2}r^{2}+2pq^{2}r-p^{2}q^{2}-p^{2}r^{2}-q^{4}-q^{2}r^{2}=0 $
Þ $ 2pq^{2}r-p^{2}r^{2}-q^{4}=0 $
Þ $ {{(q^{2}-pr)}^{2}}=0 $ Hence $ q^{2}=pr $ . Thus p, q, r in G.P.
BETA
