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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.

Parcly Taxel
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Bailey B.
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3 Answers3

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\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: $3.1875=.50(\sqrt{3.0^2+x^2})$, $6.375=\sqrt{3.0^2+x^2}$. $x≈±5.62$

AfronPie
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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 $x = \pm 5.625$.

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

Parcly Taxel
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