Question

Challenge!

If \(a, b, c \in \mathbb{R}\) and \(c(a+b+c)<0\), then the quadratic equation \(ax^2 + bx + c = 0\) definitely has:

A. A negative root.
B. Two complex roots.
C. Two real roots.
D. Two positive real roots.

Answer 1

I predict that it is D

Answer 2

Please show the solution along with your answer.

Answer 3

I assume a and b are negative and c remains positive. And (a + b) < c

This creates the scenario where c(a + b + c) < 0

Next I utilize b^2 - 4ac
Since a and b is negative
(-b)^2 - 4(-a)(c) = b^2 + 4ac which means the result will be positive

As long as a + b < c then the result will always yield positive discriminant.

Answer 4

a = b = -2;
c = 1;
We have the following roots
\[
\left\{\left\{x\to \frac{1}{2}
\left(-1-\sqrt{3}\right)\right\},\left\{x\to \frac{1}{2}
\left(\sqrt{3}-1\right)\right\}\right\}
\]

Answer 5

Positive discriminant \(\ne\) positive roots.

Answer 6

What if c is negative and a+b>|c| ?

Answer 7

Correction: The answer is C. If you have a positive discriminant, then there will always be two real roots.

Answer 8

simply f(0). f(1) < 0
use location of roots to predict its nature....
and one of the root will be negative

Answer 9

Bansal Classes. :D

By the way, what if the graph is downwards? Still negative?

Answer 10

yep...bansal
still the same...

Answer 11

I've almost given off the solution. So I'll better do the whole thing now.

[SPOILER ALERT]

@divu.mkr followed the right path.
\[c(a + b + c) < 0\]\[\Rightarrow f(0) \times f(1) < 0\]So \(f(0)\) and \(f(1)\) have the opposite sign.
So we have a root lying between \(0\) and \(1\).
For obvious reasons, the other root is real too.

And so we're left with C as @Hero said.

Answer 12

Just noticed that the question has a gap.

Answer 13

Very cool(unexpected) way to use IVT ! xD

Answer 14

Haha, yeah. I was stunned at the solution.

By the way, err, this expression can be a perfect square too.