Complex Numbers And Quadratic Equations question 700
Question: Suppose the quadratic equations $ x^{2}+px+q=0 $ and $ x^{2}+rx+s=0 $ are such that p, q, r, s are real and $ pr=2(q+s). $ Then
Options:
A) Both the equations always have real roots
B) At least one equation always has real roots
C) Both the equation always have non real roots
D) At least one equation always has real and equal roots
Show Answer
Answer:
Correct Answer: B
Solution:
Let the discriminant of the equation $ {x^{2}}+px+q=0byD_1 $ , then $ D_1=p^{2}-4q $ and the discriminant $ D_2 $ of the equation $ {x^{2}}+rx+s=0isD=r^{2}-4s $
$ \therefore D_1+D_2=p^{2}+r^{2}-4(q+s)=p^{2}+r^{2}-2pr $ [from the given relation]
$ \therefore D_1+D_2 ={{( p-r )}^{2}} \ge 0 $ Clearly at least one of $ {D_1}andD_2 $ must be non-negative consequently at least one of the equation has real roots.
BETA
