+0

# tk8 prob 6

0
762
2
+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.

Mar 13, 2019

#1
+118471
+4

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

Mar 13, 2019
#2
+257
+1

Thank you for explaining and letting me know I was wrong!

SydSu22  Mar 14, 2019