UPSKILL MATH PLUS

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### Theory:

Let us remember the formulas for solving and simplifying the radicals of positive integers $$m$$, $$n$$ and positive rational numbers $$a$$ and $$b$$.

 S. No Radical notation Index notation $$1$$ ${\left(\sqrt[n]{a}\right)}^{n}=a$ ${\left({a}^{\frac{1}{n}}\right)}^{n}=a$ $$2$$ $\sqrt[n]{a}×\sqrt[n]{b}=\sqrt[n]{\mathit{ab}}$ ${a}^{\frac{1}{n}}×{b}^{\frac{1}{n}}={\left(\mathit{ab}\right)}^{\frac{1}{n}}$ $$3$$ $\sqrt[m]{\sqrt[n]{a}}=\sqrt[\mathit{mn}]{a}=\sqrt[n]{\sqrt[m]{a}}$ ${\left({a}^{\frac{1}{n}}\right)}^{\frac{1}{m}}={a}^{\frac{1}{\mathit{mn}}}={\left({a}^{\frac{1}{m}}\right)}^{\frac{1}{n}}$ $$4$$ $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$ $\frac{{a}^{\frac{1}{n}}}{{b}^{\frac{1}{n}}}={\left(\frac{a}{b}\right)}^{\frac{1}{n}}$
Example:
Write the surd $$\sqrt{96}$$ in its simplest form.

Solution:

$$\sqrt{96} = \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 3}$$

$$= \sqrt{2^2 \times 2^2 \times 2 \times 3}$$

$$= \sqrt{2^2} \times \sqrt{2^2} \times \sqrt{2} \times \sqrt{3}$$ (Law of radicals ($$2$$))

$$= 2 \times 2 \times \sqrt{2} \times \sqrt{3}$$ (Law of radicals ($$1$$))

$$= 4 \times \sqrt{2} \times \sqrt{3}$$

Therefore, the simplest form is $$4 \times \sqrt{2} \times \sqrt{3}$$.