I know how to solve quadratic equations when there's only one unknown, but I'm a bit confused on what I do if I have $2$ unknowns e.g: $$x^2 + 2kx + 81 =0$$ With just $x^2 + 2kx + c=0$, obviously $x=-k \pm \sqrt{k^2 -c}$.
I tried substituting the above value into the equation where $c$ is $81$ and then solving it, but I just thought that I was overcomplicating everything. How do I get the possible values of $k$ in this equation when there's an $x$ too?
Remember the discriminant of the quadratic $ax^2 + bx + c = 0$ is $b^2 - 4ac$. You need the discriminant to be positive to get real solutions in $x$.
Hence in your equation you need to find for which $k$ to you get : $$(2k)^2 - 4\cdot 81 \ge 0 \qquad \Longleftrightarrow \qquad k^2 \ge 81$$
Ok to solve? Don't forget $k$ could be negative ;-)
In order to get the possible values of $k$ in the equation $$x^2 + 2kx + 81 =0$$ consider the system:
\begin{cases}
y=-\frac{x^2+81}{2x}\\
y=k
\end{cases}
The solution of $k$ would be the intersection of the two functions: the second one is simply a stack of lines parallel to $x-$axis, while the first one should be studied graphically with a sketch. Let $f(x) = -\frac{x^2+81}{2x}$, $f'(x) = \frac{81}{2x^2}-\frac12$, and $f'(x) = 0 \iff x=\pm 9\Rightarrow$ there 2 are stationary points in $x=\pm9$, so the tangents to the function $f(x)$ are parallel to $x-$axis:
Finally it can be stated that: \begin{cases} 0 \text{ solutions for }x \iff -9\lt k \lt 9\\ 1 \text{ solution for }x \iff k=\pm9\\ 0 \text{ solutions for }x \iff k\lt -9 \lor k\gt 9\\ \end{cases} So all the possible values of $k$ are in $[9;+\infty[$ and in $]-\infty;-9[$.
