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

28 Points
```
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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All the best Palak J !!!

Regards,
```
7 years ago
mycroft holmes
271 Points
```										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
```
7 years ago
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