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.
Best Answer
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: $$4(k+2)^2 -4(k+1)(k-1)>0$$ $$4(k+2)^2 -4(k+1)(k-1)=0$$ $$4(k+2)^2 -4(k+1)(k-1)<0$$