Radical Expression Practice Questions

To simplify a square root, find the largest perfect square factor of the number under the radical. The largest perfect square factor of 72 is 36:

\(\sqrt{72} = \sqrt{36 \times 2}\)

Use the property \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\) to separate the factors:

\(= \sqrt{36} \times \sqrt{2} = 6\sqrt{2}\)

Choices C and D are not fully simplified because 8 and 18 each still contain perfect square factors.

Taking the square root of a variable expression is equivalent to raising it to the power of \(\tfrac{1}{2}\). Multiply the exponents:

\(\sqrt{x^{10}} = (x^{10})^{1/2} = x^{10 \times 1/2} = x^5\)

Alternatively, you can think of it this way: since \((x^5)^2 = x^{10}\), the square root of \(x^{10}\) is \(x^5\).

When adding radical expressions, you can only combine terms that have the same radicand (the number under the radical). Both terms here have \(\sqrt{5}\), so add the coefficients:

\(3\sqrt{5} + 7\sqrt{5} = (3 + 7)\sqrt{5} = 10\sqrt{5}\)

This works just like combining like terms. Choice A incorrectly adds the radicands together (\(5 + 5 = 10\)), which is not how radical addition works.

You cannot add \(\sqrt{12}\) and \(\sqrt{27}\) directly because they have different radicands. First, simplify each radical:

\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)

\(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\)

Now both terms share the same radicand, so they can be combined:

\(2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}\)

Choice A incorrectly adds the radicands (\(\sqrt{12 + 27} = \sqrt{39}\)). Remember: \(\sqrt{a} + \sqrt{b} \neq \sqrt{a + b}\).

When multiplying square roots, use the property \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \times b}\):

\(\sqrt{3} \cdot \sqrt{15} = \sqrt{3 \times 15} = \sqrt{45}\)

Now simplify \(\sqrt{45}\) by finding its largest perfect square factor, which is 9:

\(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\)

Choice A is not fully simplified because 45 still contains a perfect square factor.

When dividing square roots, use the property \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\):

\(\dfrac{\sqrt{48}}{\sqrt{3}} = \sqrt{\dfrac{48}{3}} = \sqrt{16} = 4\)

To rationalize a denominator, multiply both the numerator and denominator by the radical in the denominator. This eliminates the radical from the denominator without changing the value of the expression:

\(\dfrac{5}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{5\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \dfrac{5\sqrt{3}}{3}\)

Choice D incorrectly removes the radical entirely from the expression instead of just from the denominator.

Break the expression into separate square roots for the coefficient and each variable:

\(\sqrt{18x^4y^6} = \sqrt{18} \cdot \sqrt{x^4} \cdot \sqrt{y^6}\)

Simplify each part individually:

\(\sqrt{18} = 3\sqrt{2}\), \(\:\sqrt{x^4} = x^2\), \(\:\sqrt{y^6} = y^3\)

Combine the results:

\(\sqrt{18x^4y^6} = 3x^2y^3\sqrt{2}\)

Choice D drops the \(\sqrt{2}\) factor, which means it is not equivalent to the original expression.

To simplify a cube root, find the largest perfect cube factor of the number under the radical. The largest perfect cube factor of 54 is 27:

\(\sqrt[\large{3}]{54} = \sqrt[\large{3}]{27 \times 2}\)

Use the property \(\sqrt[\large{3}]{a \times b} = \sqrt[3]{a} \times \sqrt[\large{3}]{b}\):

\(= \sqrt[\large{3}]{27} \times \sqrt[\large{3}]{2} = 3\sqrt[\large{3}]{2}\)

When multiplying radical expressions, multiply the coefficients together and the radicands together:

\(2\sqrt{6} \cdot 3\sqrt{10} = (2 \times 3) \cdot \sqrt{6 \times 10} = 6\sqrt{60}\)

Now simplify \(\sqrt{60}\). The largest perfect square factor of 60 is 4:

\(6\sqrt{60} = 6\sqrt{4 \times 15} = 6 \cdot 2\sqrt{15} = 12\sqrt{15}\)

Choice A is not fully simplified because 60 still contains a perfect square factor.

A rational (fractional) exponent can be converted to radical form using the rule:

\(x^{m/n} = \sqrt[\large{n}]{x^m}\)

The denominator of the fraction becomes the index of the radical, and the numerator becomes the power inside. For \(x^{2/3}\):

\(x^{2/3} = \sqrt[\large{3}]{x^2}\)

Choice D is a common error that treats the fractional exponent as division by 3 rather than a radical.

When the denominator is a binomial containing a radical, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of \(2 + \sqrt{3}\) is \(2 – \sqrt{3}\):

\(\dfrac{4}{2 + \sqrt{3}} \cdot \dfrac{2 – \sqrt{3}}{2 – \sqrt{3}}\)

Multiply the numerator:

\(4(2 – \sqrt{3}) = 8 – 4\sqrt{3}\)

Multiply the denominator using the difference of squares pattern:

\((2 + \sqrt{3})(2 – \sqrt{3}) = 4 – 3 = 1\)

Since the denominator simplifies to 1:

\(\dfrac{8 – 4\sqrt{3}}{1} = 8 – 4\sqrt{3}\)

This expression is a product of conjugates, which follows the difference of squares pattern:

\((a + b)(a – b) = a^2 – b^2\)

Here, \(a = \sqrt{5}\) and \(b = 2\):

\((\sqrt{5})^2 – (2)^2 = 5 – 4 = 1\)

This property is exactly why multiplying by the conjugate is used to rationalize denominators — it eliminates the radical.

Unlike square roots, cube roots can be taken of negative numbers. Break the expression into separate cube roots:

\(\sqrt[\large{3}]{-64x^9} = \sqrt[\large{3}]{-64} \cdot \sqrt[\large{3}]{x^9}\)

Simplify each part. Since \((-4)^3 = -64\), we have \(\sqrt[\large{3}]{-64} = -4\). For the variable, divide the exponent by 3:

\(\sqrt[\large{3}]{x^9} = x^{9/3} = x^3\)

Combine the results:

\(\sqrt[\large{3}]{-64x^9} = -4x^3\)

Divide the coefficients and the radicals separately:

\(\dfrac{6\sqrt{20}}{2\sqrt{5}} = \dfrac{6}{2} \cdot \dfrac{\sqrt{20}}{\sqrt{5}}\)

Simplify the coefficient: \(\frac{6}{2} = 3\). For the radicals, use the quotient property:

\(\dfrac{\sqrt{20}}{\sqrt{5}} = \sqrt{\dfrac{20}{5}} = \sqrt{4} = 2\)

Multiply the results:

\(3 \times 2 = 6\)