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The equation \(\frac{x}{x+1} + \frac{x}{x+2} = kx\)

has exactly two different complex solutions. Find all possible complex values for k.

Enter all the possible values, separated by commas.

 Aug 12, 2022
edited by sherwyo  Aug 12, 2022
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Hi sherwyo!

Notice, x=0 is a solution, but it is not what we want, so now suppose \(x \neq 0\), and divide both sides by x:

\(\dfrac{1}{x+1}+\dfrac{1}{x+2}=k \\ \iff \dfrac{2x+3}{(x+1)(x+2)} = k \hspace{0.5cm} \text{Common denominator} \\ \text{Next, multiply both sides by (x+1)(x+2):} \\ \iff 2x+3 = k(x+1)(x+2) \hspace{0.5cm} \text{Expand the R.H.S:} \\ \iff 2x+3=k(x^2+3x+2) \space \space \space \text{Simplify} \\ \iff 2x+3=kx^2+3kx+2k \\ \iff kx^2+(3kx-2x)+(2k-3)=0 \\ \iff kx^2+(3k-2)x+(2k-3)=0 \\ \)

Now this is a quadratic equation. For this quadratic equation to have two complex roots, the disciminant: \(\Delta =b^2-4ac <0\) (Must be less than zero.)

So:

\((3k-2)^2-4(k)(2k-3) < 0 \\ \iff (9k^2-12k+4)-8k^2+12 <0 \\ \iff k^2-12k+16 <0 \\ \text{The critical points (when the above inequality is = 0) }\space \space \space k_{1,2}=6 \pm 2\sqrt{5}\)

So the desired solution of the inequality:    \(6-2\sqrt{5}

 Aug 13, 2022

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