Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square of both sides and I tried to factor and failed.
 A: \begin{align} 
& \sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=.25 \\
& \sqrt{3.5^2+x^2}=.25+\sqrt{3.0^2+x^2} \\
\end{align}
Next, square both sides. You get:
\begin{align}
& 3.5^2+x^2=(.25+\sqrt{3.0^2+x^2})^2 \\
3.5^2+x^2=.0625+.50(\sqrt{3.0^2+x^2})+3.0^2+x^2
\end{align}
Simplifying further gets you: $3.1875=.50(\sqrt{3.0^2+x^2})$, $6.375=\sqrt{3.0^2+x^2}$. 
$x≈±5.62$ 
A: The following is a shortcut that works for this particular equation.
Multiplying both sides by the sum of the radicals (which is always $\ne 0$) gives:
$$(3.5^2+x^2)-(3.0^2+x^2)=0.25(\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2})$$
$$\sqrt{3.5^2+x^2}+\sqrt{3.0^2+x^2} = 13$$
Adding the latter to the original equation gives:
$$2 \sqrt{3.5^2+x^2} = 13.25$$
From there $x^2 = (\frac{13.25}{2})^2 - 3.5^2 = 43.890625 - 12.25 = 31.640625$, thus $x = \pm 5.625$.
A: The idea is to move the equation's terms around such that one and only one square root appears on one side. Squaring then removes that isolated square root; the other side may still have a square root, but there will be one less overall. Repeat until a polynomial is obtained.
Let $y=x^2$, then (using fractions instead of decimals) we have
$$\sqrt{\frac{49}4+y}-\sqrt{9+y}=\frac14$$
$$\sqrt{\frac{49}4+y}=\sqrt{9+y}+\frac14$$
$$\frac{49}4+y=\left(\sqrt{9+y}+\frac14\right)^2=9+y+\frac12\sqrt{9+y}+\frac1{16}$$
$$\frac{49}4-9-\frac1{16}=\frac12\sqrt{9+y}=\frac{51}{16}$$
$$\sqrt{9+y}=\frac{51}{8}$$
$$9+y=\frac{2601}{64}$$
$$y=\frac{2025}{64}$$
$$x=\pm\frac{45}8=\pm5.625$$
