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 $D>=0$
$D=b^2-4ac$
$rArr12^2-4*2*2k>=0$
$rArr144-16k>=0$
$rArr 16K<=144$
$rArr K<=9$