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?
Best Answer
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 ;-)