Originally Posted by
HallsofIvy That is, as you have been told, impossible. If you multiply both sides of $\displaystyle q/x- 2x= 3$ by x you get $\displaystyle q- 2x^2= 3x$ which is a quadratic equation and has, at most, 2 distinct roots. The original equation might have fewer solutions, but cannot have more so NO value of q gives the equation 4 roots.
If, on the other hand, you are saying that 4 is a root, we must have $\displaystyle q/4- 2(4)= 3$ so that $\displaystyle q/4= 11$ and then $\displaystyle q= 44$.
What is the exact statement of the problem in your text? You posted a picture in your first post, but that is something you wrote. Could you post a picture of the problem in the text?