What are the values of k such that 2x^2-12x+2k=0 has two solutions?

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Answer 1 · 2018-05-28T11:18:38

It must be #9>k#

Dividing your equation by #2#
#x^2-6x+k=0#
using the quadratic formula
#x_{1,2}=3pmsqrt{9-k}#
so we get two real Solutions for
#9>k#

Answer 2 · 2018-05-28T11:22:52

#K<=9#

For two Solution Discriminant(D) should be #D>=0#
#D=b^2-4ac#
#rArr12^2-4*2*2k>=0#
#rArr144-16k>=0#
#rArr 16K<=144#
#rArr K<=9#