Maximum compression of spring

[songoku]

Homework Statement

A spring is positioned vertical with the bottom end connected to ground. An object falls directly above the spring and when it almost hit the spring, it has speed of 3 m/s. Given that the mass of the object is 5 kg and the spring constant is 200 N/m, find the maximum compression of the spring

Relevant Equations

Conservation of energy
KE = 1/2 m v^2
PE = mgh
Elastic potential energy = 1/2 k x^2

1/2 m v² + mgh = 1/2 k x²

1/2 (5) (9) + (5) (9.81) x = 1/2 (200) x²

100 x² - 49.05 x - 22.5 = 0

x = 0.779 m or x = - 0.289 m

Answer key says the answer is 0.289 m but in my opinion the answer should be 0.799 m because I take h = 0 at the position where the spring has maximum compression so at the point where the object almost hit the spring the value of h should be positive, hence the value of x also should be positive.

Where is my mistake?

Thanks

 

[haruspex]

You might notice that the mid point between the two answers is the equilibrium position. The two values individually represent the extremes of the oscillation that would result if the mass were attached to the spring.
(The equation you wrote describes both positions.)
So the value with the greater magnitude, being a distance from the relaxed position of the top of the spring, must be the one representing the bottom of the oscillation.
In short, you are right, they are wrong.

 

[songoku]

Thank you very much haruspex