341Check your logic carefully. You actually have to prove that you cannot have p^2 \le 4q and r^2 \le 4s.
Now see how you can use the given condition pr = 2(q+s) to arrive at a contradiction if both inequalities are true.
if 4 real nos p, q, r & s are such that,
pr=2(q+s)
prove,
at least one of the 2 foll eqns
x2+px+q=0 and x2+rx+s=0
my approach:
i found out the discriminant of the 2 which came to be as
p2-4q and r2-4s
now one must be positive and other negative,
hence the to prove is (p2-4q)(r2-4s)<0
am i right?
and also cannot proceed after this...how to proceed?
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9 Answers
341
71at least one of the 2 foll eqns.... ? What's next. Is it incomplete!
49atgs is nickname given to me by aveek and some of the people who no longer use t iit like akari!!
1i am seeing u atgs everyday :P some people are not easily forgettable
hows ur prep going on
3is the last part of the question..."one of the foll. equations have real roots" ???
then the soultion is simple.....prove that
d1+d2 ≥ 0 usind the given relation!! thats it subhomoy!!
341pr = 2(q+s) \Rightarrow (pr)^2 = 4(q+s)^2 \ge 16 (qs)%2 as (q+s)^2 \ge 4qs
Now if both the equations have non-real roots we would have p^2 < 4q, r^2 < 4s \Rightarrow (pr)^2 < 16qs
Hence at least one of the equations has real roots