+257 The two imaginary solutions of (x-1)(x-2)(x-3)=(6-1)(6-2)(6-3) satisfy the equation x^2+k=0 What is the value of k? I got +-36 but this isn't imaginary so I don't know if this is correct.
+118687 No it is not correct. 
I think i did it the hard way,....
(x-1)(x-2)(x-3)=(6-1)(6-2)(6-3)
So x=6 is one solution.
It is a cubic so there is a total of 3 answers for x.
The other 2 satisfy x^2+k=0
Maybe there is a quicker way to do it but I expanded the left.
\((x-1)(x-2)(x-3) =5*4*3\\ x^3-6x^2+11x-6=60\\ x^3-6x^2+11x-66=0\\ \text{Then I divided the LHS by (x-6) and got}\\ (x-6)(x^2+11)=0\\ So \\ x=6,\;and\;\;x^2=-11\\ x^2=-11\\ x=\pm\sqrt{-11}\\ x=x=\pm\sqrt{11}\;i\\ \)
But you were not asked for all the x solutions you were asked for k and k=11
