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

Feb 20, 2017

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

$k \in \left(- \infty , 0\right] \cup \left[8 , \infty\right)$

#### Explanation:

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

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

is in the standard form:

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

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

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- k\right)}^{2} - 4 \left(1\right) \left(2 k\right) = {k}^{2} - 8 k = k \left(k - 8\right)$

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

Hence:

$\Delta \ge 0 \iff k \in \left(- \infty , 0\right] \cup \left[8 , \infty\right)$