Apply the formula and see where it leads.
u = a + bi + cj + dk
v = a2-b2-c2-d2 + 2a(bi+cj+dk)
v = a2 - (b′2+c′2+d′2)/a2 + 2(b′i+c′j+d′k)
Well you can tell straight away what b′ c′ and d′ are. Substitute for these and set the first expression to the real component of v. This gives a quadratic in a2. Solve the quadratic, then take the square root of that to find a. Of course a belongs to R, so the quadratic has to produce a square in R. In the reals, the quadratic has to yeild a positive solution.
Let's take the square root of 3+4i over the reals. With b′ = 2, solve a4-3a2-4 = 0. This gives a2 = 4 or -1, the latter being impossible. Thus a is ±2, and the square root is plus or minus 2+i.
Assume the real component of v is j. Now -j becomes the middle coefficient on the quadratic formula, and the discriminant is larger than j2, whence its square root is larger than j. Exactly one of the two solutions is positive, and v has exactly two square roots.