Find the set of real values k, k ≠ -1, for which the roots of the equation x^{2} + 4x - 1 + k(x^{2} + 2x + 1) = 0 are complex with positive real part.
So far, I have found the discriminant of the equation and solved for k when Δ < 1, giving me k < -5/4. But this is only the complex(?) part of the solution. How do I find the "positive real part"?
I cannot read post 2 for some reason so this post may be entirely repetitive.
First off, your logic in post 1 was perfect.
$x^2 + 4x - 1 + k(x^2 + 2x + 1) = 0 \implies (k + 1)x^2 + (2k + 4)x + (k - 1) = 0 \implies x = \dfrac{-\ (2k + 4) \pm \sqrt{(2k + 4)^2 - 4(k + 1)(k - 1)}}{2(k + 1)} \implies$
$x = \dfrac{(-2)(k + 2) \pm \sqrt{4k^2 + 16k + 16 - 4k^2 + 4}}{2(k + 1)} = \dfrac{-k - 2 \pm \sqrt{4k + 5}}{(k + 1)}.$
Now if $4k + 5 \ge 0$, then x has no imaginary part. So for x to have an imaginary part $\implies 4k + 5 < 0 \implies 4k < - 5 \implies k < -\ \dfrac{5}{4}.$ Well done.
So consider x if k < - (5/4).
$x = \dfrac{-k - 2 \pm \sqrt{(5 - 4k)(-\ 1)}}{k + 1} = \left(- \dfrac{k + 2}{k + 1}\right) \pm \left(\dfrac{\sqrt{5 - 4k}}{k + 1}\right) * i.$ You with me?
So what determines if the real part of x is positive? k + 2 and k + 1 must have opposite signs. Does that make sense?
OK. But $1 < 2 \implies k + 1 < k + 2.$ So the only way that they can have opposite signs is if k + 2 is positive and k + 1 is negative. Stop me if this is unclear.
$k + 2 > 0 \implies k > -\ 2.$ And $k + 1 < 0 \implies k < - 1 = -\ \dfrac{4}{4}.$ But we have already determined that $k < -\ \dfrac{5}{4} < -\ \dfrac{4}{4}.$
And so to meet both conditions, an imaginary part and a positive real part, $\implies -\ 2 < k < -\ \dfrac{5}{4}.$
Excuse me if this just repeats post 2, which does not render for me. Let us know if you have further questions.