First of all, let me state the problem and my rationale:
Show the range of values of $k$ for which $x^2-kx+2k+1=0$ has distinct real roots
Premise I: The discriminant is: $b^2 – 4ac$*
Premise II: $ Discriminant > 0$, otherwise if $discriminant=0$ it has only one root and if $ discriminant < 0$ then it's not in the real domain.
Then:
- $b^2 – 4ac > 0$
- $(-k)^2 -4(1 * 2k +1)$
- $k^2 – 8k – 4 > 0 $
- $(k-4)^2 – 4^2 – 4 > 0$
Completing the squares - $(k-4)^2 – 20 > 0$
- $(k-4)^2 > 20$
At this point I can tell that this is zero if $ (k – 4)^2 = 20 $
Hence, we can deduce:
$|k-4| >\sqrt{20}$
This tell us that either $k-4 > \sqrt{20}$ or that $-(k-4) > \sqrt{20}$
Therefore, the range is:
$\{k: k < 4- 2\sqrt{5}\} \cup \{k: k > 4 +2\sqrt{5}\} $
because that's when $k^2 -8k-4 > 0$
Great.
Now to the question:
Since $a > 0$ this parabola will have a minimum. Where's the vertex of this equation?
Getting the vertex form $a(x -h) + k$ on step 5 I get the coordinates $(-4,-20)$. But that can't be since $(k-4)^2>20$ so that it can have two distinct roots. Besides, getting $(-4,-20)$ yields a strangely shaped parabola that grows towards the northeast side of my Cartesian plan.
Thanks a lot in advance!
Edit-1: Replacing "k" by "discriminant" on Premise II.
Best Answer
If construct a Cartesian plane in which you label one axis $k$ and the other axis $y,$ then the graph of the equation $y = k^2−8k−4$ is a parabola with vertex $(-4,-20),$ which is what you find when you apply the vertex form $y = a(k-h)^2 + L$ to the equation $y = k^2−8k−4.$
But I think the parabola whose vertex you are supposed to find is the graph of $$y = x^2−kx+2k+1$$ in a conventional Cartesian plane where the axes are labeled $x$ (not $k$) and $y.$ So you should apply the vertex form to $x^2−kx+2k+1$ (viewed as a quadratic over $x,$ where $k$ is some as-yet-unspecified constant) rather than to $k^2−8k−4$ (viewed as a quadratic over $k$).