I have a solution - $0.1016$ meters - to a physics problem. I am told that it is incorrect. I furnish a solution, and I still think the solution is correct unless the mistake is with approximating the answer. The directions are to give the answer "to three significant digits." 

**Problem**

Two beads with a mass of $1$ kilogram are at the ends of a compressed spring resting at the same distance above the $x$-axis on a smooth parabolic wire $y = x^{2}$. The spring has a natural length of $1$ meter and a spring constant of $5$ Newtons per meter. If the system is in static equilibrium, and if the acceleration due to gravity is $10$ meters per second squared, how far away from the $y$-axis is each bead? Use three significant digits. 

**Calculation of the length of the compressed spring**

Since the natural length of the spring is $1$ meter, if the beads are stationary at $P = (x, \, x^{2})$ and $Q = (-x, \, x^{2})$ for some real number $0 < x < 1/2$, the spring exerts a force of
\begin{equation\*}
\frac{5}{2} \Bigl(1 - 2x\Bigr) \, \boldsymbol{i}
\end{equation\*}
away from the $y$-axis on each bead. If $\textbf{N}$ Newtons is the force exerted on the bead at $P$ by the wire, the magnitude of $\textbf{N}$ is equal to $\textbf{w}\_{\perp}$, the component of the weight of the beads that is normal to the parabola at $P$. The weight of each bead is $10$ Newtons. Since the slope of the tangent line at $P$ is $2x$,
\begin{equation\*}
\textbf{w}\_{\perp} = \frac{20x}{\sqrt{(2x)^{2} + 1}} \, \boldsymbol{i} + \frac{10}{\sqrt{(2x)^{2} + 1}} \, \boldsymbol{j} .
\end{equation\*}
So,
\begin{equation\*}
\frac{5}{2} \Bigl(1 - 2x\Bigr) = \frac{20x}{\sqrt{(2x)^{2} + 1}} ,
\end{equation\*}
\begin{equation\*}
1 - 2x = \frac{8x}{\sqrt{(2x)^{2} + 1}} ,
\end{equation\*}
\begin{equation\*}
4x^{2} - 4x + 1 = \frac{64x^{2}}{(2x)^{2} + 1} ,
\end{equation\*}
and
\begin{equation\*}
16x^{4} - 16x^{3} - 56x^{2} - 4x + 1 = 0 .
\end{equation\*}
Approximate solutions to this quartic equation are $-1.3805$, $-0.1811$, $0.1016$, and $2.46$. Only one of these approximations is between 0 and $1/2$: $0.1016$. So, $P$ and $Q$ are $0.1016$ meters from the $y$-axis.