Grade 10 questions on how to rationalize radical expressions are presented below. Review the formulas and examples, try solving the practice questions, and click the arrows to view the step-by-step solutions.
To rationalize radical expressions with denominators is to express the denominator without radicals.
The following identities may be used to rationalize denominators of rational expressions:
Review the following examples to see how rationalization is applied:
Rationalize: $\dfrac{1}{\sqrt{2}}$
Because of $\sqrt{2}$ in the denominator, multiply the numerator and denominator by $\sqrt{2}$ and simplify:
$$ \dfrac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{(\sqrt{2})^2} = \dfrac{\sqrt{2}}{2} $$
Rationalize: $\dfrac{1}{\sqrt[3]{x}}$
Because of $\sqrt[3]{x}$ in the denominator, multiply the numerator and denominator by $\left( \sqrt[3]{x} \right)^2$ and simplify:
$$ \dfrac{1}{\sqrt[3]{x}} = \dfrac{1}{\sqrt[3]{x}} \cdot \dfrac{\left( \sqrt[3]{x} \right)^2}{\left( \sqrt[3]{x} \right)^2} = \dfrac{\sqrt[3]{x^2}}{x} $$
Rationalize: $\dfrac{4}{\sqrt{3} - \sqrt{2}}$
Because of the expression $\sqrt{3} - \sqrt{2}$ in the denominator, multiply the numerator and denominator by its conjugate, which is $\sqrt{3} + \sqrt{2}$, to obtain:
$$ \dfrac{4}{\sqrt{3} - \sqrt{2}} = \dfrac{4}{\sqrt{3} - \sqrt{2}} \cdot \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} $$
$$ = \dfrac{4(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} $$
$$ = \dfrac{4(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^{2} - (\sqrt{2})^{2}} $$
$$ = \dfrac{4(\sqrt{3} + \sqrt{2})}{3 - 2} = 4(\sqrt{3} + \sqrt{2}) $$
Rationalize: $\dfrac{5x^{2}}{\sqrt[3]{x^{2}}}$
Because of the expression $\sqrt[3]{x^2}$ in the denominator, multiply the numerator and denominator by $\left( \sqrt[3]{x^2} \right)^2$ to obtain:
$$ \dfrac{5x^2}{\sqrt[3]{x^2}} = \dfrac{5x^2}{\sqrt[3]{x^2}} \cdot \dfrac{\left( \sqrt[3]{x^2} \right)^2}{\left( \sqrt[3]{x^2} \right)^2} $$
$$ = \dfrac{5x^2 \sqrt[3]{x^4}}{\left( \sqrt[3]{x^2} \right)^3} $$
Simplify and cancel terms:
$$ = \dfrac{5x^2 \sqrt[3]{x^4}}{x^2} = 5\sqrt[3]{x^4} = 5x\sqrt[3]{x} $$
Rationalize: $\dfrac{x^2}{y + \sqrt{x^2 + y^2}}$
Because of the expression $y + \sqrt{x^2 + y^2}$ in the denominator, multiply the numerator and denominator by its conjugate, $y - \sqrt{x^2 + y^2}$, to obtain:
$$ \dfrac{x^2}{y + \sqrt{x^2 + y^2}} = \dfrac{x^2}{y + \sqrt{x^2 + y^2}} \cdot \dfrac{y - \sqrt{x^2 + y^2}}{y - \sqrt{x^2 + y^2}} $$
$$ = \dfrac{x^2(y - \sqrt{x^2 + y^2})}{(y)^2 - (\sqrt{x^2 + y^2})^2} $$
$$ = \dfrac{x^2(y - \sqrt{x^2 + y^2})}{y^2 - (x^2 + y^2)} $$
$$ = \dfrac{x^2(y - \sqrt{x^2 + y^2})}{-x^2} $$
$$ = -y + \sqrt{x^2 + y^2} $$
Rationalize the denominators of the following expressions and simplify if possible:
Simplify: $\dfrac{10}{\sqrt{5}}$
Multiply the numerator and denominator by $\sqrt{5}$:
$$ \frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{(\sqrt{5})^2} = \frac{10\sqrt{5}}{5} = 2\sqrt{5} $$
Simplify: $2\sqrt{2}\sqrt{3} - \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}}$
Focus on the fraction part. Multiply its numerator and denominator by the conjugate, $\sqrt{2} - \sqrt{3}$:
$$ = 2\sqrt{2}\sqrt{3} - \left( \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} \right) $$
Simplify the fraction:
$$ = 2\sqrt{2}\sqrt{3} - \frac{(\sqrt{2}-\sqrt{3})^2}{(\sqrt{2})^2-(\sqrt{3})^2} $$
$$ = 2\sqrt{2}\sqrt{3} - \frac{(\sqrt{2})^2 + (\sqrt{3})^2 - 2\sqrt{2}\sqrt{3}}{2-3} $$
$$ = 2\sqrt{2}\sqrt{3} - \frac{2+3-2\sqrt{2}\sqrt{3}}{-1} $$
Distribute the negative sign:
$$ = 2\sqrt{2}\sqrt{3} + 5 - 2\sqrt{2}\sqrt{3} $$
$$ = 5 $$
Simplify: $\dfrac{7x^4}{\sqrt[3]{x^4}}$
Multiply the numerator and denominator by $\left( \sqrt[3]{x^4} \right)^2$:
$$ \frac{7x^4}{\sqrt[3]{x^4}} = \frac{7x^4}{\sqrt[3]{x^4}} \cdot \frac{\left( \sqrt[3]{x^4} \right)^2}{\left( \sqrt[3]{x^4} \right)^2} $$
Simplify:
$$ = \frac{7x^4 \sqrt[3]{x^8}}{\left( \sqrt[3]{x^4} \right)^3} = \frac{7x^4 \sqrt[3]{x^8}}{x^4} = 7 \sqrt[3]{x^8} = 7x^2 \sqrt[3]{x^2} $$
Simplify: $\dfrac{-x^2}{y + \sqrt{x^2 + y^2}}$
Multiply the numerator and denominator by the conjugate, $y - \sqrt{x^2 + y^2}$:
$$ \frac{-x^2}{y + \sqrt{x^2 + y^2}} = \frac{-x^2}{y + \sqrt{x^2 + y^2}} \cdot \frac{y - \sqrt{x^2 + y^2}}{y - \sqrt{x^2 + y^2}} $$
Simplify the denominator:
$$ = \frac{-x^2(y - \sqrt{x^2 + y^2})}{y^2 - (x^2 + y^2)} $$
$$ = \frac{-x^2(y - \sqrt{x^2 + y^2})}{-x^2} = y - \sqrt{x^2 + y^2} $$