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

Question

4131 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 (-oo, 0] uu [8, oo)#

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

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

is in the standard form:

#ax^2+bx+c = 0#

with #a=1#, #b=-k# and #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)#