Add and Subtract Radical Expressions

Detailed Questions, Explanations, and Solutions for Grade 10 Math

To add and subtract radical expressions, you must identify "like" radicals. This process is identical to combining like terms in polynomial algebra.

Need to review the basics? Start with our introduction to radicals.

Definition of Like Radicals

Radical expressions are considered like only if they possess the exact same index and the same radicand.

  • \( 6 \sqrt[3]{5} \) and \( -5 \sqrt[3]{5} \) are like (Index: 3, Radicand: 5).
  • \( 7 \sqrt[3]{8} \) and \( -5 \sqrt[3]{9} \) are not like (Radicands 8 and 9 differ).
  • \( 3 \sqrt{2x} \) and \( -5 \sqrt{2x} \) are like (Index: 2, Radicand: \(2x\)).

Basic Addition & Subtraction Examples

Only like radicals may be added or subtracted by factoring out the common radical and combining their coefficients.

Example 1: \( 4 \sqrt[3]{5} + 7 \sqrt[3]{5} \)
\[ 4\sqrt[3]{5} + 7\sqrt[3]{5} = \sqrt[3]{5}(4+7) = 11\sqrt[3]{5} \]
Example 2: \( 9 \sqrt{13} - 11 \sqrt{13} \)
\[ 9\sqrt{13} - 11\sqrt{13} = \sqrt{13}(9-11) = -2\sqrt{13} \]
Example 3: \( -8 \sqrt[4]{2x+1} + 6 \sqrt[4]{2x+1} \)
\[ \sqrt[4]{2x+1}(-8+6) = -2\sqrt[4]{2x+1} \]
Example 4: \( -\sqrt{2xy} - 4 \sqrt{2xy} + 23 \sqrt{2xy} \)
\[ \sqrt{2xy}(-1-4+23) = 18\sqrt{2xy} \]

Transforming Unlike Radicals

When radicands differ, simplify them first by factoring out perfect squares (or cubes) to see if they can be converted into like radicals.

Example 5: \( 4\sqrt{8} - 6\sqrt{2} \)
\[ 4\sqrt{2^2 \cdot 2} - 6\sqrt{2} = 4 \cdot 2\sqrt{2} - 6\sqrt{2} = 2\sqrt{2} \]
Example 6: \( -4\sqrt{12} + 12\sqrt{108} \)

Prime factorize 12 and 108: \(12 = 2^2 \cdot 3\) and \(108 = 2^2 \cdot 3^3\).

\[ -4\sqrt{2^2 \cdot 3} + 12\sqrt{2^2 \cdot 3^2 \cdot 3} \] \[ = -8\sqrt{3} + 72\sqrt{3} = 64\sqrt{3} \]
Example 7: \( \sqrt[4]{(x+1)} + 3\sqrt[4]{16(x+1)} \)
\[ \sqrt[4]{(x+1)} + 3 \cdot 2\sqrt[4]{(x+1)} = \sqrt[4]{(x+1)} + 6\sqrt[4]{(x+1)} = 7\sqrt[4]{x+1} \]

Practice Questions with Solutions

Question 1: \( -2\sqrt{3}+4\sqrt{3}+20 \)

\[ (-2+4)\sqrt{3} + 20 = 2\sqrt{3} + 20 \]

Question 2: \( 20\sqrt{7}-2\sqrt{28}-7 \)

\[ 20\sqrt{7} - 2\sqrt{4 \cdot 7} - 7 = 20\sqrt{7} - 4\sqrt{7} - 7 = 16\sqrt{7} - 7 \]

Question 3: \( -\sqrt{32}-2\sqrt{50}+3\sqrt{200} \)

Simplify: \(32 = 2 \cdot 16\), \(50 = 2 \cdot 25\), \(200 = 2 \cdot 100\).

\[ -4\sqrt{2} - 10\sqrt{2} + 30\sqrt{2} = 16\sqrt{2} \]

Question 4: \( -\sqrt{\frac{28}{9}}+3\sqrt{\frac{63}{25}} \)

\[ -\frac{2\sqrt{7}}{3} + 3 \cdot \frac{3\sqrt{7}}{5} = \sqrt{7}\left(-\frac{2}{3} + \frac{9}{5}\right) = \frac{17}{15}\sqrt{7} \]

Question 5: \( 2\sqrt{3x^2}-5\sqrt{12x^2} \)

\[ 2|x|\sqrt{3} - 5(2|x|\sqrt{3}) = -8|x|\sqrt{3} \]

Question 6: \( (7\sqrt{3x}-11\sqrt{27x})^2 \)

\[ (7\sqrt{3x} - 33\sqrt{3x})^2 = (-26\sqrt{3x})^2 = 676 \cdot 3x = 2028x \]