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I was going through my year 12 text book doing complex numbers when in chapter review I was faced with a question I've got no idea how to answer.

Consider the equation $x^2+4x-1+k(x^2+2x+1)=0$. Find the set of real values for $k$ where $k$ $\neq -1$ for which the two solutions of the equation are:
Real & Distinct, Real & Equal, Complex with positive real part and non-zero imaginary part

Please help me guys, there is nothing like this in the chapter questions and even my teacher is stumped as the book has the answers but no working out.

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  • $\begingroup$ What happens to the two roots of the polynomial when the discriminant is real, $0$ and negative? $\endgroup$ – Retty Mar 21 '17 at 23:17
  • $\begingroup$ I think you have a typo: It probably was written as "does not = -1". $\endgroup$ – quasi Mar 21 '17 at 23:18
  • $\begingroup$ Have you tried using the quadratic equation??? This is pretty trivial with it. Also, try the simple cases $x=0,1$ as a warm up (assuming you meant that k should not be $-1$, as this would create a degenerate quadratic) $\endgroup$ – Brevan Ellefsen Mar 21 '17 at 23:23
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    $\begingroup$ $(0)$ Correct the typo -- it should be $k \ne -1$.$\;(1)$ Leaving $k$ unknown, collect the terms in powers of $x$, with coefficients expressed in terms of $k$ $\;(2)$ Solve for $x$ by the quadratic formula. $\;(3)$ Look at that the expression inside the square root -- what if it's positive?; what if it's zero?; what if it's negative? $\;(4)$ Of the $3$ cases, in one case you get non-real roots, but in that case, the real part, expressed in terms of $k$, is what? $\;(5)$ So now you have some work to do. $\endgroup$ – quasi Mar 21 '17 at 23:30
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In short, you have $$(k+1)x^2+2(k+2)x+(k-1)=0$$ Hence by the quadratic formula you get: $$x=\frac{-2(k+2)\pm\sqrt{4(k+2)^2-4(k+1)(k-1)}}{2(k+1)}$$ The bit under the square root is called the $discriminant, D$. In short, because square rooting a negative number gives a complex number, it will tell you whether the roots to your equation are real $(D\ge0)$ or complex, $(D\lt0)$. The special case is when $D=0$. There, this equation simplifies to $$x=\frac{-2(k+2)\pm0}{2(k+1}$$ and both solutions are identical. What this question is asking you to find is what $k$ needs to be such that $D\gt0$ (Real and Distinct), $D=0$ (Real and Equal) and $D<0$ (complex).

So solve the following inequalities: $$4(k+2)^2 -4(k+1)(k-1)>0$$ $$4(k+2)^2 -4(k+1)(k-1)=0$$ $$4(k+2)^2 -4(k+1)(k-1)<0$$

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