[based on MAT 2018 (G)] The curves y = x^2 + c and y^2 = x touch at a single point. Find c.

If helpful draw a diagram to see what is going on, but keep the time in mind! From your sketch it should be apparent that c must be positive and that the single point of contact of the curves is above the x-axis, so that we may consider the second curve as y = x^1/2 . ( This second point is useful for (2) below. )
We are asked to find the number c, such that the two curves touch at a single point. Call this point (x_p , y_p ). For the two parabolas to touch at (x_p , y_p ) we require that (1) (x_p , y_p ) lies on both curves. Hence, y_p = x_p²+ c and y_p ²= x_p .(2) The gradients of the two curves are equal at (x_p , y_p ). Hence, 2x_p = (1/2)* x_p^-1/2 , by differentiating the equations of the curves and equating.
Now, let's start by solving (2) directly. This gives x_p= 4^-2/3 . That's all we are going to get from (2), so move onto (1). The first equation in (1) has two unknowns where as the second equation in (1) has only one unknown, y_p- start there! Use y_p ²= x_p to find y_p as y_p = 4^-1/3. Finally, the first equation in (1) allows us to solve for c ! Upon substitution of x_p= 4^-2/3 and y_p = 4^-1/3 we obtain, c = y_p- x_p² = 4^-1/3 - 4^-4/3 = 4^-1/3(1-1/4) = 3/(44^1/3) . The answer is c = 3/(44^1/3) .