Question

Write the simplest rationalization factor of the following surds:

(i) $\sqrt{32}$

(ii) $\sqrt{x^3{y}^5}$

(iii) $2\sqrt{2}$

(iv) $\frac{3}{5}\sqrt{10}$

(v) 3$\sqrt{72}$

(vi) 2$\sqrt[3]{5}$

(vii) 4 $\sqrt[3]{3}$

(viii) $\sqrt[4]{7}$

(ix) $4\sqrt[5]{9}$

(x) $\sqrt[5]{a^2{b}^3{c}^4}$

Answer 1

(i)
$$\begin{gathered} \sqrt{32}\times \sqrt{2}=\sqrt{64}=8 \\ \mathrm{Hence},\sqrt{2}\mathrm{is}\mathrm{a}\mathrm{rationali}\text{s}\mathrm{ing}\mathrm{factor}\mathrm{of}\sqrt{32}. \end{gathered}$$

(ii)
$$\begin{gathered} \sqrt{x^3{y}^5}\times \sqrt{xy}=\sqrt{x^4{y}^6}=x^2{y}^3 \\ \mathrm{Hence},\sqrt{xy}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}\sqrt{x^3{y}^5}. \end{gathered}$$

(iii)

$$\begin{gathered} 2\sqrt{2}\times \sqrt{2}=2\times 2=4 \\ \mathrm{Hence},\sqrt{2}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}2\sqrt{2}. \end{gathered}$$

(iv)

$$\begin{gathered} \frac{3}{5}\sqrt{10}\times \sqrt{10}=\frac{3}{5}\times 10=6 \\ \mathrm{Hence},\sqrt{10}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}\frac{3}{5}\sqrt{10}. \end{gathered}$$

(v)
$$\begin{gathered} 3\sqrt{72}\times \sqrt{2}=3\times 12=36 \\ \mathrm{Hence},\sqrt{2}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}3\sqrt{72}. \end{gathered}$$

(vi)
$$\begin{gathered} 2\sqrt[3]{5}\times \sqrt[3]{25}=2\times 5=10 \\ \mathrm{Hence},\sqrt[3]{25}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}2\sqrt[3]{5}. \end{gathered}$$

(vii)
$$\begin{gathered} 4\sqrt[3]{3}\times \sqrt[3]{9}=4\times 3=12 \\ \mathrm{Hence},\sqrt[3]{9}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}4\sqrt[3]{3}. \end{gathered}$$

(viii)

$$\begin{gathered} \sqrt[4]{7}\times \sqrt[4]{343}=7 \\ \mathrm{Hence},\sqrt[4]{343}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}\sqrt[4]{7}. \end{gathered}$$

(ix)
$$\begin{gathered} 4\sqrt[5]{9}\times \sqrt[5]{27}=4\times 3=12 \\ \mathrm{Hence},\sqrt[5]{27}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}4\sqrt[5]{9.} \end{gathered}$$

(x)
$$\begin{gathered} \sqrt[5]{a^2{b}^3{c}^4}\times \sqrt[5]{a^3{b}^{2}c}=abc \\ \mathrm{Hence},\sqrt[5]{a^3{b}^{2}c}\mathrm{is}\mathrm{a}\mathrm{rationali}s\mathrm{ing}\mathrm{factor}\mathrm{of}\sqrt[5]{a^2{b}^3{c}^4}. \end{gathered}$$