Add and Subtract Radical Expressions Questions with Solutions

Grade 10 questions on how to add and subtract expressions with radicals and their solutions are presented.

Definition

Radical expressions are like if they have the same index and the same radicand.
Examples
\[ 1) \; 6 \sqrt[3]{5} \quad \text{and} \quad -5 \sqrt[3]{5} \] are like radicals because they have the same index (root number which is 3) and the same radicand (number under the radical which is 5). \[ 2)\; 7 \sqrt[3]{8} \quad \text{and} \quad -5 \sqrt[3]{9} \] are not like radicals because they have different radicands 8 and 9. \[ 3)\; 3 \sqrt{2x} \quad \text{and} \quad -5 \sqrt{2x} \] are like radicals because they have the same index (2 for square root) and the same radicand \(2 x \).

Add and Subtract Like Radicals

Only like radicals may be added or subtracted.

Examples

Simplify the following expressions \[ 1)\; 4 \sqrt[3]{5} + 7 \sqrt[3]{5} \] \[2)\; 9 \sqrt{13} - 11 \sqrt{13} \] \[3)\; -8 \sqrt[4]{2x+1} + 6 \sqrt[4]{2x+1} \] \[4)\; -\sqrt{2xy} - 4 \sqrt{2xy} + 23 \sqrt{2xy} \]

Solutions to the Above Examples

The above expressions are simplified by first factoring out the like radicals and then adding/subtracting. \[ 1)\; 4\sqrt[3]{5} + 7\sqrt[3]{5} = \sqrt[3]{5}(4+7) = 11\sqrt[3]{5} \] \[2)\; 9\sqrt{13} - 11\sqrt{13} = \sqrt{13}(9-11) = -2\sqrt{13} \] \[3)\; -8\sqrt[4]{2x+1} + 6\sqrt[4]{2x+1} = \sqrt[4]{2x+1}(-8+6) = -2\sqrt[4]{2x+1} \] \[4)\; -\sqrt{2xy} - 4\sqrt{2xy} + 23\sqrt{2xy} = \sqrt{2xy}(-1-4+23) = 18\sqrt{2xy} \]

More Examples

Simplify the following expressions Here are the expressions from the image formatted in LaTeX for MathJax: \[ 1) \; 4\sqrt{8} - 6\sqrt{2} \] \[ 2) \; 5\sqrt[3]{81} - 6\sqrt[3]{3} \] \[ 3) \; -4\sqrt{12} + 12\sqrt{108} \] \[ 4) \; -\sqrt{20x} - 4\sqrt{45x} \] \[ 5) \; \sqrt[4]{(x+1)} + 3\sqrt[4]{16(x+1)} \]

Solutions to Above Examples

The above expressions are simplified by first transforming the unlike radicals to like radicals and then adding/subtracting
1) \( 4\sqrt{8} - 6\sqrt{2} = 4\sqrt{2^2 \cdot 2} - 6\sqrt{2} \) \[ = 4\sqrt{2^2} \cdot \sqrt{2} - 6\sqrt{2} \] \[ = 4 \cdot 2\sqrt{2} - 6\sqrt{2} = 2\sqrt{2} \] 2) \( 5\sqrt[3]{81} - 6\sqrt[3]{3} = 5\sqrt[3]{27 \cdot 3} - 6\sqrt[3]{3} \) \[ = 5\sqrt[3]{27} \cdot \sqrt[3]{3} - 6\sqrt[3]{3} \] \[ = 5 \cdot 3\sqrt[3]{3} - 6\sqrt[3]{3} \] \[ = 15\sqrt[3]{3} - 6\sqrt[3]{3} = 9\sqrt[3]{3} \]
3) \( -4\sqrt{12} + 12\sqrt{108} \)
When it is not obvious to obtain a common radicand from 2 different radicands, decompose them into prime numbers. Decompose 12 and 108 into prime factors as follows. \[ 12 = 2^2 \cdot 3 \quad \text{and} \quad 108 = 2^2 \cdot 3^3 \] We now substitute 12 and 108 by their prime factors and simplify: \[ -4\sqrt{12} + 12\sqrt{108} = -4\sqrt{2^2 \cdot 3} + 12\sqrt{2^2 \cdot 3^3} \] \[ = -4 \cdot \sqrt{2^2} \cdot \sqrt{3} + 12 \cdot \sqrt{2^2} \cdot \sqrt{3^3} \] \[ = -4 \cdot 2 \cdot \sqrt{3} + 12 \cdot 2 \cdot \sqrt{3^2 \cdot 3} \] \[ = -8\sqrt{3} + 24 \cdot 3\sqrt{3} \] \[ = -8\sqrt{3} + 72\sqrt{3} = 64\sqrt{3} \] 4) \( -\sqrt{20x} - 4\sqrt{45x} = -\sqrt{2^2 \cdot 5x} - 4\sqrt{3^2 \cdot 5x} \) \[ = -2\sqrt{5x} - 4 \cdot 3\sqrt{5x} = -2\sqrt{5x} - 12\sqrt{5x} = -14\sqrt{5x} \]
5) \( \sqrt[4]{(x+1)} + 3\sqrt[4]{16(x+1)} = \sqrt[4]{(x+1)} + 3\sqrt[4]{2^4(x+1)} \) \[ = \sqrt[4]{(x+1)} + 3 \cdot 2\sqrt[4]{(x+1)} = \sqrt[4]{(x+1)} + 6\sqrt[4]{(x+1)} = 7\sqrt[4]{x+1} \]

Questions With Solutions

Simplify the following expressions

\( \quad -2\sqrt{3}+4\sqrt{3}+20 \)
\( \quad 20\sqrt{7}-2\sqrt{28}-7 \)
\( \quad -\sqrt{32}-2\sqrt{50}+3\sqrt{200} \)
\( \quad 2\sqrt{4x}-3\sqrt{x} \)
\( \quad -\sqrt{\frac{28}{9}}+3\sqrt{\frac{63}{25}} \)
\( \quad 6. 2\sqrt{3x^2}-5\sqrt{12x^2} \)
\( \quad 2\sqrt[3]{40x^3}-5x\sqrt[3]{5}+\sqrt[3]{135x^3} \)
\( \quad (7\sqrt{3x}-11\sqrt{27x})^2 \)

Solutions to the Above Questions

\[ -2\sqrt{3} + 4\sqrt{3} + 20 = 2\sqrt{3} + 20 \]
\[ 20\sqrt{7} - 2\sqrt{28} - 7 = 20\sqrt{7} - 2\sqrt{4 \cdot 7} - 7 \] \[ = 20\sqrt{7} - 2 \cdot 2\sqrt{7} - 7 = 16\sqrt{7} - 7 \]
The 3 radicands in the given expression \(\sqrt{32} - 2\sqrt{50} + 3\sqrt{200}\) are different but note that 32, 50 and 200 may be written as 2 times a number that is a perfect square as follows: \(32 = 2 \cdot 16\), \(50 = 2 \cdot 25\) and \(200 = 2 \cdot 100\). Substitute in the given expression and simplify. \[ -\sqrt{32} - 2\sqrt{50} + 3\sqrt{200} = -\sqrt{2 \cdot 16} - 2\sqrt{2 \cdot 25} + 3\sqrt{2 \cdot 100} \] \[ = -4\sqrt{2} - 2 \cdot 5\sqrt{2} + 3 \cdot 10\sqrt{2} \] \[ = -4\sqrt{2} - 10\sqrt{2} + 30\sqrt{2} = 16\sqrt{2} \]
\[ 2\sqrt{4x} - 3\sqrt{x} = 2 \cdot \sqrt{4} \cdot \sqrt{x} - 3\sqrt{x} \] \[ = 2 \cdot 2\sqrt{x} - 3\sqrt{x} = 4\sqrt{x} - 3\sqrt{x} = \sqrt{x} \]
\[ -\sqrt{\frac{28}{9}} + 3\sqrt{\frac{63}{25}} = -\frac{\sqrt{28}}{\sqrt{9}} + 3\frac{\sqrt{63}}{\sqrt{25}} \] Decompose 28 and 63 into prime factors as follows: \( 28=2^2 \cdot 7 \), \( 63=3^2 \cdot 7 \) and substitute into the given expression and simplify \[ = -\frac{\sqrt{2^2 \cdot 7}}{3} + 3 \cdot \frac{\sqrt{3^2 \cdot 7}}{5} = -\frac{2\sqrt{7}}{3} + 3 \cdot \frac{3\sqrt{7}}{5} \] Factor \(\sqrt{7}\) out and simplify \[ = \sqrt{7} \left( -\frac{2}{3} + \frac{9}{5} \right) = \sqrt{7} \left( \frac{-10}{15} + \frac{27}{15} \right) = \sqrt{7} \cdot \frac{17}{15} = \frac{17}{15} \sqrt{7} \]
\[ 2\sqrt{3x^2} - 5\sqrt{12x^2} = 2\sqrt{3x^2} - 5\sqrt{4 \cdot 3x^2} \] \[ = 2\sqrt{3} \cdot |x| - 5 \cdot 2\sqrt{3} \cdot |x| = 2|x|\sqrt{3} - 10|x|\sqrt{3} = -8|x|\sqrt{3} \]
\[ 2\sqrt[3]{40x^3} - 5x\sqrt[3]{5} + \sqrt[3]{135x^3} = 2\sqrt[3]{8 \cdot 5x^3} - 5x\sqrt[3]{5} + \sqrt[3]{27 \cdot 5x^3} \] \[ = 2\sqrt[3]{2^3 \cdot 5x^3} - 5x\sqrt[3]{5} + \sqrt[3]{3^3 \cdot 5x^3} \] \[ = 2 \cdot 2x\sqrt[3]{5} - 5x\sqrt[3]{5} + 3x\sqrt[3]{5} = 4x\sqrt[3]{5} - 5x\sqrt[3]{5} + 3x\sqrt[3]{5} = 2x\sqrt[3]{5} \]
\[ (7\sqrt{3x} - 11\sqrt{27x})^2 = (7\sqrt{3x} - 11\sqrt{9 \cdot 3x})^2 \] \[ = (7\sqrt{3x} - 11 \cdot 3\sqrt{3x})^2 = (7\sqrt{3x} - 33\sqrt{3x})^2 \] \[ = (-26\sqrt{3x})^2 = (-26)^2 \cdot (\sqrt{3x})^2 = 676 \cdot 3x = 2028x \]