We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
137
2
avatar+250 

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+102786 
+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+250 
+1

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

SydSu22  Mar 14, 2019

4 Online Users