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

Question

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

Jitender Singh · Answer

Ans:
Hello Student,
Please find answer to your question below

$f(x) = 2ax^2 + 3bx + 5c$

Since this equation does not have any real roots.
So f(x) will either be positve or negative.
Lets check out.
$f(0) = 2a(0)^2 + 3b(0) + 5c$
$f(0) = 5c$
Since c > 0 (Given)
So
f(0) > 0
So f(x) is +ve for all x belongs to real.
$f(-1) = 2a(-1)^2 + 3b(-1)+5c$
$f(-1) = 2a - 3b + 5c$
f(-1) > 0
2a – 3b + 5c > 0
Hence Proved.

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

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

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

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

1 Answers

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