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avatar+256 

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
avatar+106885 
+4

No it is not correct.    sad

 

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
avatar+256 
+1

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

SydSu22  Mar 14, 2019

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