Grade 10 questions on how to simplify radical expressions are presented below. Review the rules and properties of radicals, try the practice problems, and click the arrows to view the step-by-step solutions.
In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals.
A) $\sqrt[n]{x^n} = |x|$ if $n$ is even
B) $\sqrt[n]{x^n} = x$ if $n$ is odd
Examples:
More examples on Roots of Real Numbers and Radicals.
The product formula for radicals with equal indices is given by:
More examples on how to Multiply Radical Expressions.
The division formula for radicals with equal indices is given by:
More examples on how to Divide Radical Expressions.
You may add or subtract like radicals only.
Example:
More examples on how to Add Radical Expressions.
You may rewrite expressions without radicals in the denominator (rationalization) using identities such as:
More examples on how to Rationalize Denominators of Radical Expressions.
Rationalize and simplify the given expressions:
Simplify: $\sqrt{128} \cdot \sqrt{32}$
Write 128 and 32 as products of prime factors: $128 = 2^7$, $32 = 2^5$.
$$ \sqrt{128} \cdot \sqrt{32} = \sqrt{2^7} \cdot \sqrt{2^5} = \sqrt{2^7 \cdot 2^5} $$
$$ = \sqrt{2^{12}} = \sqrt{(2^6)^2} = |2^6| = 64 $$
Simplify: $\sqrt{2} \cdot \sqrt{6} + 3\sqrt{12}$
Use the product rule to write $\sqrt{2} \cdot \sqrt{6} = \sqrt{12}$.
$$ \sqrt{2} \cdot \sqrt{6} + 3\sqrt{12} = \sqrt{2 \cdot 6} + 3\sqrt{12} = \sqrt{12} + 3\sqrt{12} $$
$$ = \sqrt{12}(1 + 3) = 4\sqrt{12} $$
$$ = 4\sqrt{4 \cdot 3} = 4\sqrt{4}\sqrt{3} = 4 \cdot 2 \cdot \sqrt{3} = 8\sqrt{3} $$
Simplify: $\dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}}$
Write 14 and 63 as products of prime numbers ($14 = 2 \times 7$, $63 = 3^2 \times 7$) and substitute:
$$ \dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}} = \dfrac{3\sqrt{2 \cdot 7} + 4\sqrt{3^2 \cdot 7}}{3\sqrt{7}} $$
$$ = \dfrac{3\sqrt{2}\sqrt{7} + 4 \cdot 3\sqrt{7}}{3\sqrt{7}} $$
$$ = \dfrac{\sqrt{7}(3\sqrt{2} + 12)}{3\sqrt{7}} $$
$$ = \dfrac{3\sqrt{2} + 12}{3} = \sqrt{2} + 4 $$
Simplify: $\sqrt[3]{32} \sqrt[3]{16}$
Write 32 and 16 as products of prime numbers ($32 = 2^5$, $16 = 2^4$) and substitute:
$$ \sqrt[3]{32} \sqrt[3]{16} = \sqrt[3]{2^5} \sqrt[3]{2^4} = \sqrt[3]{2^5 \cdot 2^4} $$
$$ = \sqrt[3]{2^9} = \sqrt[3]{(2^3)^3} = 2^3 = 8 $$
Simplify: $\sqrt[3]{\dfrac{64}{7}}$
Write 64 as a product of prime numbers ($64 = 2^6$) and substitute:
$$ \sqrt[3]{\dfrac{64}{7}} = \sqrt[3]{\dfrac{2^6}{7}} = \sqrt[3]{\dfrac{(2^2)^3}{7}} = \dfrac{4}{\sqrt[3]{7}} $$
Rationalize the denominator by multiplying the numerator and denominator by $\left(\sqrt[3]{7}\right)^2$:
$$ = \dfrac{4}{\sqrt[3]{7}} \cdot \dfrac{\left(\sqrt[3]{7}\right)^2}{\left(\sqrt[3]{7}\right)^2} = \dfrac{4\left(\sqrt[3]{7}\right)^2}{7} = \dfrac{4\sqrt[3]{49}}{7} $$
Simplify: $\dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}}$
Write 54 as a product of prime numbers ($54 = 2 \times 3^3$) and substitute:
$$ \dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2 \cdot 3^3 x}}{\sqrt[3]{2}} $$
$$ = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2x} \cdot \sqrt[3]{3^3}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - 3\sqrt[3]{2x}}{\sqrt[3]{2}} $$
$$ = \dfrac{-2\sqrt[3]{2x}}{\sqrt[3]{2}} = -2\sqrt[3]{\dfrac{2x}{2}} = -2\sqrt[3]{x} $$
Simplify: $\dfrac{\sqrt{x} + \sqrt{x + 1}}{\sqrt{x} - \sqrt{x + 1}}$
Multiply the denominator and numerator by the conjugate of the denominator:
$$ \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} - \sqrt{x+1}} \cdot \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+1}} $$
Expand and simplify:
$$ = \dfrac{(\sqrt{x})^2 + (\sqrt{x+1})^2 + 2\sqrt{x}\sqrt{x+1}}{(\sqrt{x})^2 - (\sqrt{x+1})^2} $$
$$ = \dfrac{x + (x + 1) + 2\sqrt{x}\sqrt{x+1}}{x - (x + 1)} $$
$$ = \dfrac{2x + 1 + 2\sqrt{x(x+1)}}{-1} $$
$$ = -2x - 1 - 2\sqrt{x(x+1)} $$
Use all the rules and properties of radicals to rationalize and simplify the following expressions.
Simplify: $\sqrt[3]{25} \cdot \sqrt[3]{125}$
$$ \sqrt[3]{25} \cdot \sqrt[3]{125} = \sqrt[3]{3125} = \sqrt[3]{125 \cdot 25} = 5\sqrt[3]{25} $$
Simplify: $(5\sqrt[3]{64})(-3\sqrt[3]{16})$
First, simplify the known cube roots ($\sqrt[3]{64} = 4$):
$$ (5 \cdot 4)(-3\sqrt[3]{16}) = 20(-3\sqrt[3]{16}) = -60\sqrt[3]{16} $$
Simplify $\sqrt[3]{16}$ as $\sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}$:
$$ -60(2\sqrt[3]{2}) = -120\sqrt[3]{2} $$
Simplify: $(7\sqrt{\dfrac{2}{5}})(2\sqrt{10})$
$$ (7\sqrt{\dfrac{2}{5}})(2\sqrt{10}) = 14\sqrt{\dfrac{20}{5}} $$
$$ = 14\sqrt{4} = 14 \cdot 2 = 28 $$
Simplify: $\sqrt[3]{2\dfrac{10}{27}}$
Convert the mixed number to an improper fraction:
$$ \sqrt[3]{2\dfrac{10}{27}} = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3} $$
Simplify: $(\sqrt{17x})(\sqrt{34x})$
$$ (\sqrt{17x})(\sqrt{34x}) = \sqrt{578x^2} = x\sqrt{578} $$
$$ = x\sqrt{289 \cdot 2} = 17x\sqrt{2} $$
(Assuming $x \ge 0$ as required by the original expression).
Simplify: $\sqrt{4y^2}$
$$ \sqrt{4y^2} = 2|y| $$
Simplify: $\sqrt{8y^4}$
$$ \sqrt{8y^4} = \sqrt{4 \cdot 2 \cdot (y^2)^2} = 2y^2\sqrt{2} $$
Simplify: $\sqrt{25+144}$
$$ \sqrt{25+144} = \sqrt{169} = 13 $$
Simplify: $\sqrt[2n]{x^{2n}}$, where $n$ is a positive integer
Because $2n$ represents an even index:
$$ \sqrt[2n]{x^{2n}} = |x| $$
Simplify: $(\sqrt{x-2})(4\sqrt{x-2})$
$$ (\sqrt{x-2})(4\sqrt{x-2}) = 4(x-2) $$
Simplify: $\dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}}$
Combine using the quotient rule for radicals:
$$ \dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}} = \sqrt[3]{\dfrac{27a^3b^5}{8a^6b^2}} $$
Simplify the fraction inside the radical:
$$ = \sqrt[3]{\dfrac{27b^3}{8a^3}} $$
Extract the perfect cubes:
$$ = \dfrac{3b}{2a} $$
Simplify: $\dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}}$
Multiply the numerator and denominator by the conjugate of the denominator, $\sqrt{x}-\sqrt{x-2}$:
$$ \dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}} = \dfrac{(\sqrt{x}-\sqrt{x-2})^2}{(\sqrt{x})^2 - (\sqrt{x-2})^2} $$
Expand the numerator and simplify the denominator:
$$ = \dfrac{x + (x-2) - 2\sqrt{x(x-2)}}{x - (x-2)} $$
$$ = \dfrac{2x - 2 - 2\sqrt{x(x-2)}}{2} $$
Divide all terms in the numerator by 2:
$$ = x - 1 - \sqrt{x(x-2)} $$