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How would I combine and simplify the following radical:

$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}$$

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3 Answers 3

$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}\\=\frac{|A|}{\sqrt 2}-\frac{|A|}{2\sqrt 2}\\=\frac{2|A|-|A|}{2\sqrt 2}\\=\frac{|A|}{2\sqrt 2}\frac{\sqrt 2}{\sqrt 2}\\=\frac{\sqrt 2|A|}{4}$$

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$\sqrt{\frac{A^2}{2}} - \sqrt{\frac{A^2}{8}} = \frac{A}{\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{2A}{2\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{A}{2\sqrt{2}}$

Assuming $A \geq 0$. If $A < 0$, you can replace with $|A|$.

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Recall the following facts. $$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} \text{ whenever }a,b>0$$ $$\sqrt{x^2} = \lvert x \rvert \text{, where $x \in \mathbb{R}$}$$ $$\sqrt{ab} = \sqrt{a} \sqrt{b} \text{ whenever }a,b>0$$ Using the above, we then get that \begin{align} \sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{8}} & = \sqrt{\dfrac{A^2}{2}} - \sqrt{\dfrac{A^2}{2^2 \times 2}} = \dfrac{\sqrt{A^2}}{\sqrt{2}} - \dfrac{\sqrt{A^2}}{\sqrt{2^2 \times2}}\\ & = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{\sqrt{2^2} \times \sqrt{2}} = \dfrac{\lvert A\rvert}{\sqrt{2}} - \dfrac{\lvert A \rvert}{2\sqrt{2}}\\ & =\dfrac{\lvert A \rvert}{2\sqrt{2}} \end{align}

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