# Competition Question with Imaginary Roots

• Sep 28th 2013, 02:08 PM
ReneG
Competition Question with Imaginary Roots
Came across this problem on a math competition.

The equation $3x^2 + 4x + k = 0$ has two imaginary roots. Which of the following is a true statement about the value of $k$?

a) $k < 0$
b) $k < -2$
c) $k > 2$
d) $k > 0$

I'm having a hard time setting up a condition for k.
• Sep 28th 2013, 02:27 PM
Plato
Re: Competition Question with Imaginary Roots
Quote:

Originally Posted by ReneG
Came across this problem on a math competition.
The equation $3x^2 + 4x + k = 0$ has two imaginary roots. Which of the following is a true statement about the value of $k$?
a) $k < 0$
b) $k < -2$
c) $k > 2$
d) $k > 0$
I'm having a hard time setting up a condition for k.

I am puzzled by this question because the correct answer is $k>\frac{4}{3}~.$

Use the discriminate, $b^2-4ac<0$ or $16-12k<0$.

The closest answer is c). But that is not the exact answer.
• Sep 28th 2013, 03:56 PM
Re: Competition Question with Imaginary Roots
I'm going to guess and say that the problem stated somewhere that k was an integer
• Sep 28th 2013, 04:02 PM
HallsofIvy
Re: Competition Question with Imaginary Roots
Surely the problem is not exactly as given here. The equation $3x^2+ 4x+ 4k= 0$ can have complex roots but cannot have imaginary roots.
• Sep 28th 2013, 04:22 PM
Plato
Re: Competition Question with Imaginary Roots
Quote:

Originally Posted by HallsofIvy
Surely the problem is not exactly as given here. The equation $3x^2+ 4x+ 4k= 0$ can have complex roots but cannot have imaginary roots.

Whereas you are absolutely correct, I was part of the mathematics contest community long enough to know many in the mathematics education community use imaginary number and complex number interchangeably. I assure you it is futile to even try to correct them. Thus having seen that this is an old contest question, I assumed the worst.