Find range of values for 'k' that gives this equation 2 distinct real roots? I stuck on a question which is asking me to find the range of values for k.
the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots.
$$3x^2+kx+2=0$$
this is what i have done so far:
$$b^2-4ac>0$$
Note - since there are two distinct real roots
$$k^2-24>0$$
considering $$k^2-24=0$$
$$k = +-\sqrt{24}$$
$$k = 2*\sqrt{6} $$or 
$$k = -2*\sqrt{6}$$
my question is how do I work out to range of values for k which makes the inequality true.
Can you solve the inequality step-by-step $$k^2-24>0$$
 A: You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > \sqrt{24} = 2\sqrt{6}$$
recalling that $\sqrt{x^2} = |x|$ ($x \in \Bbb R$). We can then write this as
$$k > 2\sqrt{6} \ \text{or} \ k < -2\sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_\pm$, say, then you just need to test the intervals $(-\infty, k_-)$, $(k_-,k_+)$ and $(k_+,\infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k \in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_\pm = \pm 2\sqrt{6}$.)
