# Math Help - Discriminants, roots and Polynomials (quick question)

1. ## Discriminants, roots and Polynomials (quick question)

Hi, we've been studying polynomials at college recently and we've covered a lot of topics on it, but I'm stuck on one question. I remember doing questions like it, but we've gone over so much in the meantime that I've forgotten how to work it out!

Find the set of values of k such that the equation $x^2 + 4kx + 3k = 0$ has two distinct roots.

The discriminant (4k squared - 4(1)(3k)) has to be > 0 for it to have 2 distinct roots, but I forgot how to work it out!

Any tips? Thanks =D

2. So you need to solve the inequality
(4k)^2-4(3k) >0

16k^2-12k>0
4k(4k-3) >0

Can you solve this inequality (so the answer will be in terms of interval)

3. Yeah, I can get to $4k(4k - 3) > 0$ but I forgot what to do next =/

I know that once I remember the method, I can work it out, but my mind's gone blank...

4. Is it $k > 3/4$ ?

Because don't you make each of the brackets equal 0 to find the intervals, so the brackets (4k)(4k - 3) would mean k would be > 0 and > 3/4

I think I'm totally off, but I tried k as 0.74 and it didn't give a positive number, but 0.76 does, so it seems right...

5. $4k \ (4k - 3) > 0$

$\Rightarrow 4k < 0$ AND $4k - 3 > 0$

$\Rightarrow k < 0$ AND $k > \frac {3}{4}$

Apparently, $k$ cannot be both $< 0$ AND $> \frac {3}{4}$

OR

$4k \ (4k - 3) > 0$

$\Rightarrow 4k > 0$ AND $4k - 3 < 0$

$\Rightarrow k > 0$ AND $k < \frac {3}{4}$

Combining these two inequalities, you get $0 < k < \frac {3}{4}.$

I hope that helps.

me07.