Worked example 1 · Easy
Worked method
Simplify \(\sqrt{75}\).
\(75=25\times3\). Therefore \(\sqrt{75}=\sqrt{25}\sqrt3=\boxed{5\sqrt3}\).
Chapter 18
Surds
Exact roots and rational denominators. Original explanations and examination-focused practice based on the supplied textbook's topic sequence.
By the end of this chapter, you should be able to
- Explain the key ideas and use correct mathematical terminology in each textbook subtopic.
- Use the appropriate methods and show enough working for examination marks.
- Check intermediate arithmetic, signs, units and the final answer.
1
Subtopic 1
Simplifying surds
A surd is an irrational root left in exact form, such as \(\sqrt2\) or \(3\sqrt5\). A perfect square has a rational square root, so \(\sqrt{49}=7\) is not a surd.
Simplify \(\sqrt n\) by finding the largest square-number factor of \(n\). Then use \(\sqrt{ab}=\sqrt a\sqrt b\). You may also combine surds under one radical when multiplying or dividing positive radicands.
Exact surd answers are preferred unless the question asks for a decimal approximation. Never claim \(\sqrt{a+b}=\sqrt a+\sqrt b\): that is generally false.
Rules and important ideas
- \(\sqrt{ab}=\sqrt a\sqrt b\quad(a,b\ge0)\)
- \(\frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab}\quad(b>0)\)
Worked example 2 · Medium
Worked method
Simplify \(\sqrt{48}\times\sqrt3\).
\(\sqrt{48}\sqrt3=\sqrt{144}=\boxed{12}\).
Worked example 3 · Exam-style
Worked method
Simplify exactly \(\sqrt{200}\div\sqrt8\).
\(\sqrt{200}\div\sqrt8=\sqrt{25}=\boxed{5}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
2
Subtopic 2
Manipulation of expressions of the form \(a+b\sqrt c\)
Expressions of the form \(a+b\sqrt c\) can be handled using normal algebra. Add or subtract the rational parts and the like surd parts separately.
Multiply brackets using distribution. In particular, \((a+b\sqrt c)(a-b\sqrt c)=a^2-b^2c\); the surd terms cancel. Square a bracket carefully: \((a+b\sqrt c)^2=a^2+2ab\sqrt c+b^2c\).
After simplification, like surds may combine—for example \(2\sqrt3+5\sqrt3=7\sqrt3\)—but unlike surds cannot be added into one term.
Rules and important ideas
- \((a+b\sqrt c)^2=a^2+2ab\sqrt c+b^2c\)
- \((a+b\sqrt c)(a-b\sqrt c)=a^2-b^2c\)
Worked example 4 · Easy
Worked method
Simplify \((5+2\sqrt3)+(4-\sqrt3)\).
Collect parts: \((5+4)+(2\sqrt3-\sqrt3)=\boxed{9+\sqrt3}\).
Worked example 5 · Medium
Worked method
Expand \((3+\sqrt2)^2\).
\(9+6\sqrt2+2=\boxed{11+6\sqrt2}\).
Worked example 6 · Exam-style
Worked method
Let \(p=4+3\sqrt5\) and \(q=4-3\sqrt5\). Find \(pq\) exactly.
Conjugate product: \(pq=4^2-(3\sqrt5)^2=16-45=\boxed{-29}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
3
Subtopic 3
Rationalising denominators
Rationalising the denominator means rewriting a fraction so no surd remains in its denominator. Multiplying the top and bottom by the same nonzero surd leaves the value unchanged.
For a denominator containing a sum or difference, multiply by its conjugate. For example, \(a+\sqrt b\) pairs with \(a-\sqrt b\). Their product is rational because the middle terms cancel.
Check the sign and simplify any common factors at the end. Do not rationalise only the denominator: multiply both numerator and denominator.
Rules and important ideas
- \(\frac1{\sqrt a}=\frac{\sqrt a}{a}\quad(a>0)\)
- \(\frac{u}{a+\sqrt b}\times\frac{a-\sqrt b}{a-\sqrt b}\)
Worked example 7 · Easy
Worked method
Rationalise \(\dfrac5{\sqrt3}\).
\(\frac5{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\boxed{\frac{5\sqrt3}{3}}\).
Worked example 8 · Medium
Worked method
Rationalise \(\dfrac4{3+\sqrt5}\).
\(\frac4{3+\sqrt5}\frac{3-\sqrt5}{3-\sqrt5}=\frac{4(3-\sqrt5)}{9-5}=\boxed{3-\sqrt5}\).
Worked example 9 · Exam-style
Worked method
Rationalise and simplify \(\dfrac{2+\sqrt3}{2-\sqrt3}\).
\(\frac{2+\sqrt3}{2-\sqrt3}\cdot\frac{2+\sqrt3}{2+\sqrt3}=\frac{(2+\sqrt3)^2}{4-3}=\boxed{7+4\sqrt3}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
Chapter 18 practice
These are newly written practice questions, not past-paper reproductions. Questions progress from confidence-building to exam-style. Answers are hidden until you open each question’s Show Answer section.
Simplifying surds
1Confidence practice[2 marks]
Simplify \(\sqrt{32}\).
Show Answer
Step-by-step solution: \(\sqrt{32}=\sqrt{16\times2}=\boxed{4\sqrt2}\).
2Confidence practice[2 marks]
Is \(\sqrt{81}\) a surd? Explain.
Show Answer
Step-by-step solution: No: \(\sqrt{81}=9\), which is rational.
3Developing practice[3 marks]
Simplify \(\sqrt{98}\).
Show Answer
Step-by-step solution: \(98=49\times2\); \(\sqrt{98}=\boxed{7\sqrt2}\).
4Developing practice[3 marks]
Calculate exactly \(\sqrt{12}\times\sqrt{27}\).
Show Answer
Step-by-step solution: \(\sqrt{324}=\boxed{18}\).
5Exam-style practice[4 marks]
Simplify \(\dfrac{\sqrt{180}}{\sqrt5}\).
Show Answer
Step-by-step solution: \(\sqrt{180/5}=\sqrt{36}=\boxed{6}\).
6Exam-style practice[4 marks]
Express \(3\sqrt{50}-\sqrt8\) in simplest surd form.
Show Answer
Step-by-step solution: \(3(5\sqrt2)-2\sqrt2=15\sqrt2-2\sqrt2=\boxed{13\sqrt2}\).
Manipulation of expressions of the form \(a+b\sqrt c\)
7Confidence practice[2 marks]
Simplify \(5\sqrt7+2\sqrt7\).
Show Answer
Step-by-step solution: \((5+2)\sqrt7=\boxed{7\sqrt7}\).
8Confidence practice[2 marks]
Simplify \((7+\sqrt3)-(2+4\sqrt3)\).
Show Answer
Step-by-step solution: \(7-2+\sqrt3-4\sqrt3=\boxed{5-3\sqrt3}\).
9Developing practice[3 marks]
Expand \((2+\sqrt5)(3-\sqrt5)\).
Show Answer
Step-by-step solution: \(6-2\sqrt5+3\sqrt5-5=\boxed{1+\sqrt5}\).
10Developing practice[3 marks]
Expand \((4-\sqrt3)^2\).
Show Answer
Step-by-step solution: \(16-8\sqrt3+3=\boxed{19-8\sqrt3}\).
11Exam-style practice[4 marks]
If \(x=5+2\sqrt7\) and \(y=5-2\sqrt7\), calculate \(xy\).
Show Answer
Step-by-step solution: \(xy=25-(2\sqrt7)^2=25-28=\boxed{-3}\).
12Exam-style practice[4 marks]
Simplify \((\sqrt3+2)^2-(\sqrt3-2)^2\).
Show Answer
Step-by-step solution: First square: \(7+4\sqrt3\); second: \(7-4\sqrt3\). Subtract: \(\boxed{8\sqrt3}\).
Rationalising denominators
13Confidence practice[2 marks]
Rationalise \(1/\sqrt2\).
Show Answer
Step-by-step solution: \(\frac1{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\boxed{\frac{\sqrt2}{2}}\).
14Confidence practice[2 marks]
Rationalise \(\dfrac7{\sqrt7}\).
Show Answer
Step-by-step solution: \(\frac7{\sqrt7}\times\frac{\sqrt7}{\sqrt7}=\frac{7\sqrt7}{7}=\boxed{\sqrt7}\).
15Developing practice[3 marks]
Rationalise \(\dfrac3{2\sqrt3}\).
Show Answer
Step-by-step solution: \(\frac3{2\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{3\sqrt3}{6}=\boxed{\frac{\sqrt3}{2}}\).
16Developing practice[3 marks]
Rationalise \(\dfrac5{4+\sqrt6}\).
Show Answer
Step-by-step solution: \(\frac5{4+\sqrt6}\cdot\frac{4-\sqrt6}{4-\sqrt6}=\frac{5(4-\sqrt6)}{16-6}=\boxed{\frac{4-\sqrt6}{2}}\).
17Exam-style practice[4 marks]
Write \(\dfrac3{5-2\sqrt2}\) in the form \(a+b\sqrt2\) with rational \(a,b\).
Show Answer
Step-by-step solution: \(\frac3{5-2\sqrt2}\cdot\frac{5+2\sqrt2}{5+2\sqrt2}=\frac{15+6\sqrt2}{25-8}=\boxed{\frac{15}{17}+\frac6{17}\sqrt2}\).
18Exam-style practice[4 marks]
Rationalise and simplify \(\dfrac{1+\sqrt3}{1-\sqrt3}\).
Show Answer
Step-by-step solution: \(\frac{(1+\sqrt3)^2}{1-3}=\frac{4+2\sqrt3}{-2}=\boxed{-2-\sqrt3}\).
High-grade revision checklist
- I can explain each rule in the exact textbook topic order.
- I can select and apply the correct method, including signs, powers and brackets.
- I can show clear working in harder multi-step questions.
- I can check by reversing the operation, estimating, or substituting a value, as appropriate.