<s>[INST] 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.
 [/INST] 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
$$\sqrt{\frac{49}4+y}-\sqrt{9+y}=\frac14$$
$$\sqrt{\frac{49}4+y}=\sqrt{9+y}+\frac14$$
$$\frac{49}4+y=\left(\sqrt{9+y}+\frac14\right)^2=9+y+\frac12\sqrt{9+y}+\frac1{16}$$
$$\frac{49}4-9-\frac1{16}=\frac12\sqrt{9+y}=\frac{51}{16}$$
$$\sqrt{9+y}=\frac{51}{8}$$
$$9+y=\frac{2601}{64}$$
$$y=\frac{2025}{64}$$
$$x=\pm\frac{45}8=\pm5.625$$
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