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

##### 1 Answer

Feb 20, 2017

#### Answer:

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

#k in (-oo, 0] uu [8, oo)#

#### Explanation:

I will assume that the question is asking for what range of values of

#x^2-kx+2k=0#

is in the standard form:

#ax^2+bx+c = 0#

with

This has discriminant

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

The graph of

Hence:

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