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Number - Surds, roots and radicals, including simplifying

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A surd is an irrational number that is expressed as a root of a rational number, such as 2\sqrt{2}, 53\sqrt[3]{5}, or 232\sqrt{3}.

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Simplifying surds involves finding the largest perfect square factor of the radicand (the number inside the root) and taking its square root outside.

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Two or more surds are 'like surds' if they have the same value under the radical symbol after simplification. Only like surds can be added or subtracted, e.g., 35+25=553\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}.

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Surds can be multiplied or divided regardless of whether they are like terms, using the laws of radicals.

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Rationalizing the denominator is the process of removing a radical from the bottom of a fraction. For a term like ka\frac{k}{\sqrt{a}}, multiply the numerator and denominator by a\sqrt{a}.

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To rationalize a binomial denominator like a+ba + \sqrt{b}, multiply both the numerator and denominator by its conjugate, a−ba - \sqrt{b}.

📐Formulae

ab=a×b\sqrt{ab} = \sqrt{a} \times \sqrt{b}

ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

ac±bc=(a±b)ca\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}

(a)2=a(\sqrt{a})^2 = a

ca=caa\frac{c}{\sqrt{a}} = \frac{c\sqrt{a}}{a}

1a+b=a−ba2−b\frac{1}{a + \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}

💡Examples

Problem 1:

Simplify the expression 75−12\sqrt{75} - \sqrt{12}.

Solution:

25×3−4×3=53−23=33\sqrt{25 \times 3} - \sqrt{4 \times 3} = 5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3}

Explanation:

Identify the largest square factors for both numbers: 2525 for 7575 and 44 for 1212. Extract the square roots and subtract the coefficients of the resulting like surds.

Problem 2:

Expand and simplify (3+2)(3−2)(3 + \sqrt{2})(3 - \sqrt{2}).

Solution:

32−(2)2=9−2=73^2 - (\sqrt{2})^2 = 9 - 2 = 7

Explanation:

This follows the difference of squares identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2. Here a=3a = 3 and b=2b = \sqrt{2}.

Problem 3:

Rationalize the denominator of 63\frac{6}{\sqrt{3}}.

Solution:

63×33=633=23\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}

Explanation:

Multiply both the numerator and denominator by 3\sqrt{3} to make the denominator a rational number (33). Then simplify the fraction 6/36/3 to 22.

Problem 4:

Simplify 18×2\sqrt{18} \times \sqrt{2}.

Solution:

18×2=36=6\sqrt{18 \times 2} = \sqrt{36} = 6

Explanation:

Use the rule a×b=ab\sqrt{a} \times \sqrt{b} = \sqrt{ab}. Since 3636 is a perfect square, the final result is a rational integer.