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
63
6
avatar

The equation \(\frac{x}{x+1} + \frac{x}{x+2} = kx\) has exactly two complex roots. Find all possible complex values for \(k.\) Enter all the possible values, separated by commas.

 May 7, 2019
 #1
avatar+568 
+1

We can start by taking the 2 smaller fractions and turning them into one bigger fraction! (I'm sorry if you don't like fractions, but making fractions makes everything a whole lot easier.) Like this: \(\frac{x(x+2)}{(x+1)(x+2)}+\frac{x(x+1)}{(x+1)(x+2)}\). This doesn't seem like it gets us anywhere but now we can make only one fraction. \(\frac{x([x+1]+[x+2])}{x^2+3x+2}=kx\) Before we divide by x, we have to solve for x=0. (I'm going to skip all of the math, you end up with 0 = 0, so there is nothing to be gained.) Now we can divide by x. \(\frac{x+1+x+2}{x^2+3x+2}\) We can simplify this: \(\frac{2x+3}{(x+1)(x+2)}\).

 

Honestly, from here, I'm stumped. Hey, CPhill, Pavlov, or Omi, could you finish this one? 

 May 9, 2019
 #2
avatar+100814 
+2

Hi Helperid, 

This question confuses me too.

Maybe another mathematician can help both of us.

 

Here is what I have done

 

The equation \(\frac{x}{x+1} + \frac{x}{x+2} = kx\) has exactly two complex roots. Find all possible complex values for \(k\)   Enter all the possible values, separated by commas.

 

x is not 0,-1 or -2

 

Divide both sides by x

 

\(\frac{x}{x+1} + \frac{x}{x+2} = kx\\ \frac{1}{x+1} + \frac{1}{x+2} = k\\ x+2+x+1=k(x+1)(x+2)\\ 2x+3=k(x^2+3x+2)\\ 2x+3=kx^2+3kx+2k\\ 0=kx^2+(3k-2)x+(2k-3)\\ \text{The roots are complex so the discriminant must be less than 0}\\ (3k-2)^2-4k(2k-3)<0\\ 9k^2-12k+4-8k^2+12k<0\\ k^2+4<0\\ k^2<-4\\ k<-2i,\quad or \quad k>2i \)

 

I don't know if this last bit is really correct anyway, what does k<-2i even mean??   PLUS

I am confused because I am told to list all possible values of k so the end is definitely not quite right.

 May 10, 2019
 #3
avatar+100549 
+1

I did this one a couple of days ago and got the same, Melody......the answers didn't make sense to me, either.....but.....maybe there is something else to be considered????

 

cool cool cool

CPhill  May 10, 2019
edited by CPhill  May 10, 2019
 #4
avatar+100814 
+1

I'd like someone with a very good understanding of complex numbers to look at this question and answer.

Maybe Heureka or Alan or Rom or Tiggsy.

Melody  May 11, 2019
 #5
avatar+5088 
+1

I think you got the answer Melody.

 

Mathematica returns

 

\(x = \dfrac{(-3 k+2) \pm \sqrt{k^2+4}}{2 k}\)

 

\(x \not \in \mathbb{R} \Rightarrow k^2 + 4 < 0\\ k^2 < -4\\ k \in i ((-\infty, 2)\cup (2, \infty))\)

 

which is a region on the imaginary axis.

Rom  May 16, 2019
edited by Rom  May 18, 2019
 #6
avatar+100814 
0

Thanks very much Rom  laugh

Melody  May 16, 2019

17 Online Users

avatar
avatar
avatar
avatar