Velocity of Mass on Spring

[KindofSlow]

Problem section is - Kinetic Energy: Work-Energy Principle

A mass "m" is attached to a spring which is held stretched a distance "x" horizontally by a force "F", and then released.
The spring compresses, pulling the mass. Assuming there is no friction, determine the speed of the mass "m" when the spring returns to half its orignal extension (x/2).

My work:
1/2*m*(v^2) = 1/2*k*((x/2)^2), also since F = k*x, then k = F/x
m*(v^2) = F/x*((x^2)/4)
v^2 = (F*x)/(4*m)
v = ((F*x)/(4*m))^(1/2)

The book says ((3*F*x)/(4*m))^(1/2)
The only difference is the 3 in the numerator under the radical.
I cannot figure out where the 3 comes from so if anyone can tell me where the 3 comes from, will greatly appreciate it.

Thank you much.

 

[KindofSlow]

If anyone cares, here is the solution.

In error, I set KE = remaining PE.
The correct equation is to set KE = PE used (not remaining).

Once KE is set equal to reduction of PE instead of PE remaining, solution easily follows and book answer is correct.

Thanks for listening.