# Finding the constant for a quadratic. Two methods; which one is correct and why?

The question reads $kx^2 + (k+2)x - 3 = 0$ has roots which are real and positive. Find the possible values k might have.

Now, since it has real and positive roots, the discriminant $\Delta{d} > 0$, so:

$(k+2)^2 - 4(k)(-3) > 0$

=> $k^2 + 4 + 4k + 12k > 0$

=> $k^2 + 16k + 4 > 0$

Now, solving it with the formula, we get:

$\frac{-16 \pm {\sqrt{250}}}{2}$

But when we factorise it, we get:

=> $k(k+16) > -4$

=> $k > -4$ or $k > -20$

BUT, that's NOT the answer, to get the answer, we have to:

$k^2 + 16k + 4 + 64> 64$

=> $k^2 + 16k + 64> 60$

=> $(k+8)^2 > 60$

=> $k > -8 + \sqrt{60}$

My question is how is this possible that I am getting three different answers and why is the last one correct?

• Why $k(k+16)$ implies $k>-4$ or $k>-20$? Moreover, $\Delta d>0$ only imply that the quadratic has solution(s). It is not enough to show that solution is positive.
– GAVD
Sep 7, 2015 at 10:47
• See the answer to this question. Sep 7, 2015 at 10:51
• Also when you are using the quadratic formula, you got something wrong. It should be 240 but not 250. $16^2 - 4(1)(4)=256-16=240$ Sep 7, 2015 at 10:51
• I totally agree with user3313320 and GAVD.That's were your mistakes lie.Correct it and you are there with your desired answer. Sep 7, 2015 at 10:56
• I am so damn stupid, really sorry to waste your time. :( Thanks a lot, regardless! :) Sep 7, 2015 at 11:06

$f(0)<0$ if $k>0$, f(x) has positive root. if $k<0$, need $-\frac{k+2}{k}>0$ →$k>-2$. So the answer is $$k>-2, (k≠0)$$