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.

Guest May 7, 2019

#1**+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?

helperid1839321 May 9, 2019

#2**+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.

Melody May 10, 2019