Exercise from Sullivan's Algebra & Trigonometry book: Chapter 1.3) Exercise 113...
Find $k$ such that the equation $kx^2 + x + k = 0$ has repeated real solutions.
I've tried to pass the $+k$ to the right and complete the square, using the quadratic formula, and grouping... the most I've reached to solve it was this expression $(\sqrt kx+\frac{1}{2\sqrt k})^2=\frac{1}{4k}-k$. But don't really know I even have to do anything like that.
If a quadratic has a repeated root, then both its roots must be the same, in which case its discriminant is equal to zero. The discriminant of the quadratic equation $$ax^2 + bx + c = 0, a \neq 0$$ is $$\Delta = b^2 - 4ac$$ Can you take it from there?