Divide Radicals Questions with Solutions for Grade 10
Grade 10 questions on how to divide radical expressions with solutions are presented.
Quotient (Division) of Radicals With the Same Index
Division formula of radicals with equal indices is given by
\[
\large {\frac{\sqrt[n]{x}}{\sqrt[n]{y}} = \sqrt[n]{\frac{x}{y}}}
\]
Examples
Simplify the given expressions
1)
\[
\frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4.
\]
2)
\[
\begin{aligned}
\quad \frac{\sqrt[3]{54x^5}}{\sqrt[3]{2x^2}} &= \sqrt[3]{\frac{54x^5}{2x^2}} \\
&= \sqrt[3]{\frac{54}{2} \cdot \frac{x^5}{x^2}} = \sqrt[3]{27x^{5-2}} \\
&= \sqrt[3]{27} \sqrt[3]{x^3} = 3x
\end{aligned}
\]
3)
\[
\begin{aligned}
\quad \frac{\sqrt{x^2 - 1}}{\sqrt{x + 1}} &= \sqrt{\frac{x^2 - 1}{x + 1}} \\
&= \sqrt{\frac{(x - 1)(x + 1)}{x + 1}} = \sqrt{x - 1} \\[10pt]
\end{aligned}
\]
4)
\[
\begin{aligned}
\quad \frac{\sqrt[3]{24y^5x^2}}{\sqrt[3]{3y^2x^5}} &= \sqrt[3]{\frac{24y^5x^2}{3y^2x^5}} \\
&= \sqrt[3]{\frac{24}{3} \cdot \frac{y^5}{y^2} \cdot \frac{x^2}{x^5}} = \sqrt[3]{8 y^{5-2} x^{2-5}} \\
&= \sqrt[3]{8 y^3 x^{-3}} = \sqrt[3]{8 \frac{y^3}{x^3}} = 2\frac{y}{x}
\end{aligned}
\]
Questions With Answers
Use the above division formula to simplify the following expressions
\[
\begin{aligned}
1. \quad &\frac{\sqrt[3]{128}}{\sqrt[3]{2}} \\[8pt]
2. \quad &\frac{\sqrt{162x^5}}{\sqrt{2x}} \\[8pt]
3. \quad &\frac{\sqrt{x+2}}{\sqrt{x^2-4}} \\[8pt]
4. \quad &\frac{\sqrt[4]{32z^6y^3}}{\sqrt[4]{2z^2y^{11}}}
\end{aligned}
\]
Solutions to the Above Problems
\[
\begin{aligned}
1. \quad \frac{\sqrt[3]{128}}{\sqrt[3]{2}} &= \sqrt[3]{\frac{128}{2}} = \sqrt[3]{64} \\
&= \sqrt[3]{2^6} = \sqrt[3]{(2^2)^3} = 2^2 = 4 \\[12pt]
2. \quad \frac{\sqrt{162x^5}}{\sqrt{2x}} &= \sqrt{\frac{162x^5}{2x}} \\
&= \sqrt{\frac{162}{2} \cdot \frac{x^5}{x}} = \sqrt{81x^4} = 9x^2 \\[12pt]
3. \quad \frac{\sqrt{x+2}}{\sqrt{x^2-4}} &= \sqrt{\frac{x+2}{x^2-4}} \\
&= \sqrt{\frac{x+2}{(x-2)(x+2)}} = \sqrt{\frac{1}{x-2}} = \frac{1}{\sqrt{x-2}} \\[12pt]
4. \quad \frac{\sqrt[4]{32z^6y^3}}{\sqrt[4]{2z^2y^{11}}} &= \sqrt[4]{\frac{32z^6y^3}{2z^2y^{11}}} \\
&= \sqrt[4]{\frac{32}{2} \cdot \frac{z^6}{z^2} \cdot \frac{y^3}{y^{11}}} = \sqrt[4]{16 \cdot \frac{z^4}{y^8}} = 2 \cdot \frac{z}{y^2}
\end{aligned}
\]
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