For what vaues of $k$ $k$ does $x^{2} - k x + 2 k = 0$ $x^2-kx+2k=0$ have a solution?

Question

4420 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T07:15:38

The given quadratic has at least one Real root if and only if:

$k \in (- \infty, 0] \cup [8, \infty)$ $k in (-oo, 0] uu [8, oo)$

I will assume that the question is asking for what range of values of $k$ $k$ does the quadratic have a Real root.

$x^{2} - k x + 2 k = 0$ $x^2-kx+2k=0$

is in the standard form:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

with $a = 1$ $a=1$ , $b = - k$ $b=-k$ and $c = 2 k$ $c=2k$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-k)^2-4(1)(2k) = k^2-8k = k(k-8)$

The graph of $k(k-8)$ is an upright parabola, with zeros at $k=0$ and $k=8$ .

Hence:

$Delta >=0 <=> k in (-oo, 0] uu [8, oo)$

For what vaues of kkk does x^2-kx+2k=0x2−kx+2k=0x^2-kx+2k=0 have a solution?