# Imaginary roots

1. Oct 27, 2014

### erisedk

1. The problem statement, all variables and given/known data
If both roots of the equation ax^2 + x + c - a = 0 are imaginary and c > -1, then:

Ans: 3a < 2+4c

2. Relevant equations
Discriminant < 0 for img roots
Vieta

3. The attempt at a solution
1-4(a)(c-a)<0
4ac > 4a^2 + 1
Minimum value of 4a^2 + 1 is 1 so
4ac>1

I can't think of anything beyond this and I have no idea how to incorporate c>-1 into my equations.

2. Oct 27, 2014

### HallsofIvy

Staff Emeritus
You mean "both roots are complex" not "imaginary", don't you? It is impossible for that equation to have imaginary roots.

3. Oct 27, 2014

### PeroK

Okay, so far. What sort of inequality is that in relation to a?

... although, I don't see how you get that answer and c > 1 would make more sense. Are you sure you've got the question right?

Last edited: Oct 27, 2014
4. Oct 27, 2014

### erisedk

Both roots are complex is the same as both roots are imaginary, I think.

The question is correct, at least should be since it's a question from a test I took yesterday.
I don't have any idea what sort of inequality that is in relation to a.

5. Oct 27, 2014

### PeroK

A complex number has real and imaginary parts. If both roots were imaginary, they would be $\pm bi$ with no real part.

Here's how I would prove the question is wrong:

Let c = 0 (which is > -1), so we have:

$ax^2 + x - a = 0$

The discriminant is $4a^2+1$ which is +ve, so this equation never has complex roots. And the condition 3a < 2 (=2 + 4c) is not relevant.

Ergo the question is wrong.

6. Oct 27, 2014

### erisedk

Wait, why is the condition 3a<2 not relevant?

7. Oct 27, 2014

### PeroK

The answer says that the equation has complex roots when 3a < 2 + 4c. In this case, when 3a < 2 (as c = 0).

But, with c = 0 the equation never has complex roots. Whether 3a < 2 or not has nothing to do with it.

As an exercise, try c = 1. You should be able to work out when the equation has complex roots and show that it not when 3a < 6

8. Oct 28, 2014

### erisedk

The equation always has complex roots for c=1.
That was neat. Thank you!