O Level (Sec 3 & 4)Additional MathematicsSingapore Syllabus
Surds: Master Surds for O Level Additional Math
Master surds with this comprehensive study guide. Learn how to simplify surds, perform operations with surds, rationalize denominators, and solve equations involving surds. Includes 7 practice questions with step-by-step solutions.
By Timothy Gan
What Are Surds?
A surd is an irrational number that cannot be expressed as a simple fraction or terminating decimal. Surds are square roots (or other roots) of numbers that are not perfect squares (or perfect powers). For example, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are surds because they cannot be simplified to rational numbers.
When we write $\sqrt{a}$, where $a$ is a positive number that is not a perfect square, we are dealing with a surd. Surds are exact values, unlike decimal approximations. For instance, $\sqrt{2} = 1.414213...$ (decimal approximation), but as a surd, we keep it in the form $\sqrt{2}$ for exactness.
Common examples of surds: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\sqrt{7}$, $\sqrt{11}$, $2\sqrt{3}$, $5\sqrt{2}$
Numbers that are NOT surds: $\sqrt{4} = 2$, $\sqrt{9} = 3$, $\sqrt{16} = 4$ (these are perfect squares)
Simplifying Surds
When simplifying surds, we look for perfect square factors. A surd is in its simplest form when the number under the square root has no perfect square factors other than 1.
To simplify a surd:
1. Find the largest perfect square factor of the number under the square root
2. Express the surd as a product of two square roots
3. Simplify the perfect square
For example, to simplify $\sqrt{12}$:
$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$
General rule: $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
Examples:
$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
$\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}$
Operations with Surds: Addition and Subtraction
When adding or subtracting surds, we can only combine like surds (surds with the same number under the square root). This is similar to collecting like terms in algebra.
Rule: $a\sqrt{b} + c\sqrt{b} = (a+c)\sqrt{b}$
Examples:
$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$
$7\sqrt{3} - 2\sqrt{3} = 5\sqrt{3}$
$4\sqrt{5} + 2\sqrt{5} - \sqrt{5} = 5\sqrt{5}$
Important: We CANNOT add or subtract surds with different numbers under the square root directly.
For example: $\sqrt{2} + \sqrt{3}$ cannot be simplified further (they are unlike surds)
However, sometimes we need to simplify surds first before we can add or subtract:
$\sqrt{8} + \sqrt{18} = \sqrt{4 \times 2} + \sqrt{9 \times 2} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$
$\sqrt{50} - \sqrt{32} + \sqrt{18} = 5\sqrt{2} - 4\sqrt{2} + 3\sqrt{2} = 4\sqrt{2}$
Operations with Surds: Multiplication
Multiplying surds is straightforward using the rule:
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
Examples:
$\sqrt{2} \times \sqrt{3} = \sqrt{6}$
$\sqrt{5} \times \sqrt{7} = \sqrt{35}$
$3\sqrt{2} \times 4\sqrt{3} = 12\sqrt{6}$
$(2\sqrt{5})(3\sqrt{2}) = 6\sqrt{10}$
When multiplying a surd by itself:
$\sqrt{a} \times \sqrt{a} = \sqrt{a^2} = a$
Examples:
$\sqrt{3} \times \sqrt{3} = 3$
$2\sqrt{5} \times 3\sqrt{5} = 6 \times 5 = 30$
Expanding brackets with surds:
$(a + \sqrt{b})(c + \sqrt{d})$ - use FOIL method (First, Outer, Inner, Last)
Example:
$(2 + \sqrt{3})(5 + \sqrt{3})$
$= 10 + 2\sqrt{3} + 5\sqrt{3} + \sqrt{3} \times \sqrt{3}$
$= 10 + 7\sqrt{3} + 3$
$= 13 + 7\sqrt{3}$
Operations with Surds: Division
When dividing surds, we use the rule:
$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
Examples:
$\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$
$\frac{\sqrt{18}}{\sqrt{3}} = \sqrt{\frac{18}{3}} = \sqrt{6}$
$\frac{6\sqrt{10}}{2\sqrt{5}} = \frac{6}{2} \times \frac{\sqrt{10}}{\sqrt{5}} = 3\sqrt{2}$
Alternatively, we can simplify by canceling common factors:
$\frac{\sqrt{50}}{\sqrt{2}} = \frac{\sqrt{25 \times 2}}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2}} = 5$
Rationalizing the Denominator
Rationalizing the denominator means eliminating surds from the denominator of a fraction. This is a standard practice in mathematics to express answers in a more acceptable form.
Case 1: Simple Surd in Denominator
To rationalize $\frac{a}{\sqrt{b}}$, multiply both numerator and denominator by $\sqrt{b}$:
$\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$
Example:
$\frac{3}{\sqrt{2}} = \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$
Case 2: Binomial with Surd in Denominator
To rationalize $\frac{a}{b + \sqrt{c}}$, multiply by the conjugate $\frac{b - \sqrt{c}}{b - \sqrt{c}}$:
The conjugate of $(a + \sqrt{b})$ is $(a - \sqrt{b})$
The conjugate of $(a - \sqrt{b})$ is $(a + \sqrt{b})$
When we multiply conjugates: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$
Example:
$\frac{1}{3 + \sqrt{2}} = \frac{1}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}}$
$= \frac{3 - \sqrt{2}}{9 - 2}$
$= \frac{3 - \sqrt{2}}{7}$
Comparing and Ordering Surds
To compare or order surds, we have several methods:
Method 1: Square both surds and compare
If $a > 0$ and $b > 0$, then:
- If $a^2 > b^2$, then $\sqrt{a} > \sqrt{b}$
- If $a^2 < b^2$, then $\sqrt{a} < \sqrt{b}$
Example: Compare $3\sqrt{2}$ and $2\sqrt{5}$
$(3\sqrt{2})^2 = 9 \times 2 = 18$
$(2\sqrt{5})^2 = 4 \times 5 = 20$
Since $18 < 20$, we have $3\sqrt{2} < 2\sqrt{5}$
Method 2: Express all surds with the same coefficient
Convert all surds to have a coefficient of 1, then compare the values under the square root.
Method 3: Convert to decimal approximations (less preferred for exact work)
Solving Equations Involving Surds
To solve equations involving surds, we typically:
1. Isolate the surd term on one side
2. Square both sides to eliminate the square root
3. Solve the resulting equation
4. Check all solutions in the original equation (squaring can introduce extraneous solutions)
Example: Solve $\sqrt{x + 5} = 7$
Step 1: Surd is already isolated
Step 2: Square both sides: $(\sqrt{x + 5})^2 = 7^2$
$x + 5 = 49$
Step 3: Solve: $x = 44$
Step 4: Check: $\sqrt{44 + 5} = \sqrt{49} = 7$ ✓
For equations with two surd terms:
Example: Solve $\sqrt{x + 1} = \sqrt{2x - 5}$
Square both sides: $x + 1 = 2x - 5$
Solve: $x = 6$
Check: $\sqrt{6 + 1} = \sqrt{7}$ and $\sqrt{2(6) - 5} = \sqrt{7}$ ✓
Warning: Always check your solutions! Squaring both sides can introduce solutions that don't work in the original equation.
Practice Questions with Solutions
Work through comprehensive solutions for each question to master surds concepts. Some questions include video explanations. Perfect for O Level Additional Math exam preparation in Singapore.
Question 1Video Solution
Simplifying Surds
Question
Simplify each of the following surds without using a calculator. (i) $\sqrt{40}$ (ii) $\sqrt{18}$ (iii) $\sqrt{396}$
Video Solution
Step-by-Step Solution
- (i) $\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}$
- (ii) $\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$
- (iii) $\sqrt{396} = \sqrt{4 \times 99} = \sqrt{4} \times \sqrt{99} = 2\sqrt{99} = 2\sqrt{9 \times 11} = 2 \times 3\sqrt{11} = 6\sqrt{11}$
Answer:(i) $2\sqrt{10}$, (ii) $3\sqrt{2}$, (iii) $6\sqrt{11}$
Question 2Video Solution
Addition and Subtraction of Surds
Question
Simplify each of the following surds without using a calculator. (i) $\sqrt{75} + \sqrt{108}$ (ii) $\sqrt{32} + \sqrt{50}$ (iii) $(3 + 5\sqrt{2})(4 - \sqrt{2})$
Video Solution
Step-by-Step Solution
- (i) $\sqrt{75} + \sqrt{108} = \sqrt{25 \times 3} + \sqrt{36 \times 3} = 5\sqrt{3} + 6\sqrt{3} = 11\sqrt{3}$
- (ii) $\sqrt{32} + \sqrt{50} = \sqrt{16 \times 2} + \sqrt{25 \times 2} = 4\sqrt{2} + 5\sqrt{2} = 9\sqrt{2}$
- (iii) $(3 + 5\sqrt{2})(4 - \sqrt{2}) = 12 - 3\sqrt{2} + 20\sqrt{2} - 5(\sqrt{2})^2 = 12 + 17\sqrt{2} - 10 = 2 + 17\sqrt{2}$
Answer:(i) $11\sqrt{3}$, (ii) $9\sqrt{2}$, (iii) $2 + 17\sqrt{2}$
Question 3Video Solution
Rationalizing the Denominator
Question
Simplify each of the following surds without using a calculator. (i) $\frac{2}{\sqrt{5}-3}$ (ii) $\frac{10}{\sqrt{2}-5}$ (iii) $\frac{8}{3\sqrt{7}-1}$
Video Solution
Step-by-Step Solution
- (i) $\frac{2}{\sqrt{5}-3} = \frac{2}{\sqrt{5}-3} \times \frac{\sqrt{5}+3}{\sqrt{5}+3} = \frac{2(\sqrt{5}+3)}{(\sqrt{5})^2 - 9} = \frac{2(\sqrt{5}+3)}{5-9} = \frac{2(\sqrt{5}+3)}{-4} = -\frac{\sqrt{5}+3}{2}$
- (ii) $\frac{10}{\sqrt{2}-5} = \frac{10}{\sqrt{2}-5} \times \frac{\sqrt{2}+5}{\sqrt{2}+5} = \frac{10(\sqrt{2}+5)}{(\sqrt{2})^2 - 25} = \frac{10(\sqrt{2}+5)}{2-25} = \frac{10(\sqrt{2}+5)}{-23} = -\frac{10(\sqrt{2}+5)}{23}$
- (iii) $\frac{8}{3\sqrt{7}-1} = \frac{8}{3\sqrt{7}-1} \times \frac{3\sqrt{7}+1}{3\sqrt{7}+1} = \frac{8(3\sqrt{7}+1)}{(3\sqrt{7})^2 - 1} = \frac{8(3\sqrt{7}+1)}{63-1} = \frac{8(3\sqrt{7}+1)}{62} = \frac{4(3\sqrt{7}+1)}{31} = \frac{12\sqrt{7}+4}{31}$
Answer:(i) $-\frac{\sqrt{5}+3}{2}$, (ii) $-\frac{10(\sqrt{2}+5)}{23}$, (iii) $\frac{12\sqrt{7}+4}{31}$
Question 4Video Solution
Simultaneous Equations with Surds
Question
It is given that $a$ and $b$ are positive integers such that $(a\sqrt{5} - 1)(\sqrt{5} + b) = 20\sqrt{5} + 32$. Form a pair of simultaneous equations and solve them to find the value of $a$ and of $b$.
Video Solution
Step-by-Step Solution
- Expand the left side: $(a\sqrt{5} - 1)(\sqrt{5} + b) = a\sqrt{5} \cdot \sqrt{5} + a\sqrt{5} \cdot b - 1 \cdot \sqrt{5} - 1 \cdot b$
- $= 5a + ab\sqrt{5} - \sqrt{5} - b = 5a - b + (ab - 1)\sqrt{5}$
- Equate to the right side: $5a - b + (ab - 1)\sqrt{5} = 20\sqrt{5} + 32$
- Compare constant terms: $5a - b = 32$ ... (1)
- Compare coefficients of $\sqrt{5}$: $ab - 1 = 20$, so $ab = 21$ ... (2)
- From equation (2), possible integer pairs $(a, b)$: $(1, 21)$, $(3, 7)$, $(7, 3)$, $(21, 1)$
- Test in equation (1): For $(a, b) = (7, 3)$: $5(7) - 3 = 35 - 3 = 32$ ✓
- Therefore, $a = 7$ and $b = 3$
Answer:$a = 7$, $b = 3$
Question 5Video Solution
Solving Equations with Surds
Question
Solve $\sqrt{7x - 5} - x - 1 = 0$.
Video Solution
Step-by-Step Solution
- Rearrange to isolate the surd: $\sqrt{7x - 5} = x + 1$
- Square both sides: $(\sqrt{7x - 5})^2 = (x + 1)^2$
- $7x - 5 = x^2 + 2x + 1$
- Rearrange to standard form: $x^2 + 2x + 1 - 7x + 5 = 0$
- $x^2 - 5x + 6 = 0$
- Factorize: $(x - 2)(x - 3) = 0$
- Therefore $x = 2$ or $x = 3$
- Check $x = 2$: $\sqrt{7(2) - 5} - 2 - 1 = \sqrt{9} - 3 = 3 - 3 = 0$ ✓
- Check $x = 3$: $\sqrt{7(3) - 5} - 3 - 1 = \sqrt{16} - 4 = 4 - 4 = 0$ ✓
Answer:$x = 2$ or $x = 3$
Question 6Video Solution
Application: Volume Problem with Surds
Question
A rectangular block has a square base. The length of each side of the base is $(\sqrt{5} - \sqrt{3})$ m and the volume of the block is $(4\sqrt{3} - 2\sqrt{5})$ m³. Find, without the use of a calculator, the height of the block in the form $a\sqrt{3} + b\sqrt{5}$.
Video Solution
Step-by-Step Solution
- Volume of rectangular block = Base area × Height
- Base area = $(\sqrt{5} - \sqrt{3})^2 = (\sqrt{5})^2 - 2\sqrt{5}\sqrt{3} + (\sqrt{3})^2 = 5 - 2\sqrt{15} + 3 = 8 - 2\sqrt{15}$
- Height = $\frac{\text{Volume}}{\text{Base area}} = \frac{4\sqrt{3} - 2\sqrt{5}}{8 - 2\sqrt{15}}$
- Simplify: $\frac{4\sqrt{3} - 2\sqrt{5}}{8 - 2\sqrt{15}} = \frac{2(2\sqrt{3} - \sqrt{5})}{2(4 - \sqrt{15})} = \frac{2\sqrt{3} - \sqrt{5}}{4 - \sqrt{15}}$
- Rationalize by multiplying by conjugate: $\frac{2\sqrt{3} - \sqrt{5}}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}}$
- Numerator: $(2\sqrt{3} - \sqrt{5})(4 + \sqrt{15}) = 8\sqrt{3} + 2\sqrt{3}\sqrt{15} - 4\sqrt{5} - \sqrt{5}\sqrt{15}$
- $= 8\sqrt{3} + 2\sqrt{45} - 4\sqrt{5} - \sqrt{75} = 8\sqrt{3} + 6\sqrt{5} - 4\sqrt{5} - 5\sqrt{3} = 3\sqrt{3} + 2\sqrt{5}$
- Denominator: $(4 - \sqrt{15})(4 + \sqrt{15}) = 16 - 15 = 1$
- Therefore, height = $3\sqrt{3} + 2\sqrt{5}$ m
Answer:$3\sqrt{3} + 2\sqrt{5}$ m (so $a = 3$, $b = 2$)
Question 7Video Solution
Application: Cylinder Volume with Surds
Question
A cylinder has a radius of $(\sqrt{2} - 1)$ cm and a volume of $(12 + 4\sqrt{2})\pi$ cm³. Find, without using a calculator, the exact value of its height, $h$, in the form $\frac{a + b\sqrt{2}}{c}$ cm, where $a$, $b$ and $c$ are integers.
Video Solution
Step-by-Step Solution
- Volume of cylinder = $\pi r^2 h$
- Given: $\pi r^2 h = (12 + 4\sqrt{2})\pi$
- Therefore: $r^2 h = 12 + 4\sqrt{2}$
- Find $r^2$: $r^2 = (\sqrt{2} - 1)^2 = (\sqrt{2})^2 - 2(\sqrt{2})(1) + 1^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$
- Height: $h = \frac{12 + 4\sqrt{2}}{r^2} = \frac{12 + 4\sqrt{2}}{3 - 2\sqrt{2}}$
- Rationalize by multiplying by conjugate: $h = \frac{12 + 4\sqrt{2}}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}}$
- Numerator: $(12 + 4\sqrt{2})(3 + 2\sqrt{2}) = 36 + 24\sqrt{2} + 12\sqrt{2} + 8(\sqrt{2})^2 = 36 + 36\sqrt{2} + 16 = 52 + 36\sqrt{2}$
- Denominator: $(3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 9 - 8 = 1$
- Therefore: $h = \frac{52 + 36\sqrt{2}}{1} = 52 + 36\sqrt{2}$ cm
- In the required form: $h = \frac{52 + 36\sqrt{2}}{1}$ cm, where $a = 52$, $b = 36$, $c = 1$
Answer:$h = \frac{52 + 36\sqrt{2}}{1}$ cm or $h = 52 + 36\sqrt{2}$ cm
Key Formulas to Remember for Surds
Product Rule for Surds
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$Note: The square root of a product equals the product of square roots
Quotient Rule for Surds
$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$Note: The square root of a quotient equals the quotient of square roots
Simplifying Surds
$\sqrt{a^2 \times b} = a\sqrt{b}$Note: Extract perfect square factors from under the square root
Adding Like Surds
$a\sqrt{b} + c\sqrt{b} = (a+c)\sqrt{b}$Note: Only surds with the same radicand can be added or subtracted
Multiplying a Surd by Itself
$\sqrt{a} \times \sqrt{a} = a$Note: Squaring a square root gives the original number
Rationalizing Simple Denominator
$\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$Note: Multiply numerator and denominator by the surd
Rationalizing Binomial Denominator
$\frac{1}{a + \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}$Note: Multiply by the conjugate to eliminate the surd
Difference of Two Squares
$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$Note: Useful when rationalizing denominators with binomials
Special Identities to Memorize
$\sqrt{a} \times \sqrt{a} = a$$\sqrt{a^2} = |a| = a$ (for $a \geq 0$)$(\sqrt{a})^2 = a$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$
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