if c is positive and 2ax²+3bx+5c=0 does not have any real roots,then prove that 2a-3b+5c>0.

Thanks!

Dear Palak J

Ans: discriminant part is >0 and hence (3b)^2 < 4*a*c and hence 3b>2(10a*c)^(0.5)

now 2a-3b+5c > 2a-2(10a*c)^(0.5)+5c

or 2a-3b+5c>(p+q)^2

where p=(2a)^(0.5)   & q=(5c)^(0.5)

As this is a square term and hence is>0

hence 2a-3b+5c>0(proved)

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Askiitians Experts
Soumyajit Das IIT Kharagpur

Since the given quadratic has no real roots, either f(x)>0 for all x or f(x)<0 for all x.

We are given that 5c = f(0)>0

Hence 2a-3b+5c = f(-1)>0