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}
