Chapter 09 — Powers (Part 4): Further Applications of Surds and Rationalisation

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MATHEMATICS HIPHOP • A-LEVEL PURE MATHEMATICS

CHAPTER 09: POWERS

PART 4: FURTHER APPLICATIONS OF SURDS AND RATIONALISATION

Advanced manipulation, exact values and algebraic applications.

Learning Objectives

By the end of this lesson, you should be able to:

• Rationalise denominators containing two different surds.
• Simplify algebraic fractions involving surds.
• Combine fractions using conjugates.
• Solve equations containing surd coefficients.
• Use surds in exact-value problems.
• Construct and understand simple proofs involving surds.
• Apply surds to geometrical and measurement problems.

1. Revision: The Conjugate Method

In Part 3, we introduced conjugates. A conjugate is formed by changing the sign between two terms.

For example:

• The conjugate of 4 + √7 is 4 − √7.
• The conjugate of √5 − √2 is √5 + √2.

The method works because the product of two conjugates is a difference of two squares.

Key identity

(a + b)(a − b) = a² − b²

The middle terms cancel, leaving a rational expression when the squares are rational.

2. Rationalising Denominators Containing Two Surds

Sometimes a denominator contains two different square roots. In this case, multiply the numerator and denominator by the conjugate of the denominator.

Worked Example 1

Rationalise the denominator of 5/(√7 + √3).

Step 1: Identify the conjugate.

The conjugate of √7 + √3 is √7 − √3.

Step 2: Multiply top and bottom by the conjugate.

= 5(√7 − √3)/[(√7 + √3)(√7 − √3)]

Step 3: Simplify the denominator.

= 5(√7 − √3)/(7 − 3)

= 5(√7 − √3)/4

Worked Example 2

Rationalise 3/(√11 − √5).

The conjugate is √11 + √5.

= 3(√11 + √5)/[(√11 − √5)(√11 + √5)]

= 3(√11 + √5)/(11 − 5)

= (√11 + √5)/2

3. Rationalising More Complicated Denominators

A denominator may contain a coefficient multiplying a surd. The same conjugate method still applies.

Worked Example 3

Rationalise 4/(3 + 2√2).

The conjugate is 3 − 2√2.

= 4(3 − 2√2)/[(3 + 2√2)(3 − 2√2)]

= 4(3 − 2√2)/(9 − 8)

= 12 − 8√2

Worked Example 4

Rationalise 6/(5 − √6).

The conjugate is 5 + √6.

= 6(5 + √6)/[(5 − √6)(5 + √6)]

= 6(5 + √6)/(25 − 6)

= (30 + 6√6)/19

4. Fractions with Surds in Both Numerator and Denominator

When both the numerator and denominator contain surds, multiply by the conjugate of the denominator and simplify the numerator carefully.

Worked Example 5

Simplify (2 + √3)/(2 − √3).

Multiply the numerator and denominator by 2 + √3:

= (2 + √3)²/[(2 − √3)(2 + √3)]

Numerator:

(2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3

Denominator:

4 − 3 = 1

Answer: 7 + 4√3

Worked Example 6

Express (3 + √2)/(3 − √2) in the form a + b√2.

Multiply by the conjugate 3 + √2:

= (3 + √2)²/[(3 − √2)(3 + √2)]

Numerator:

= 9 + 6√2 + 2

= 11 + 6√2

Denominator:

= 9 − 2 = 7

Answer: 11/7 + (6/7)√2

Therefore a = 11/7 and b = 6/7.

5. Adding and Subtracting Fractions Involving Surds

Fractions involving surds can be combined using a common denominator. Conjugates can sometimes make the process much simpler.

Worked Example 7

Simplify 1/(2 − √3) + 1/(2 + √3).

Take the common denominator:

= [(2 + √3) + (2 − √3)]/[(2 − √3)(2 + √3)]

Numerator:

= 2 + √3 + 2 − √3 = 4

Denominator:

= 4 − 3 = 1

Answer: 4

Worked Example 8

Simplify 1/(√5 − √2) − 1/(√5 + √2).

Take the common denominator:

= [(√5 + √2) − (√5 − √2)]/[(√5 − √2)(√5 + √2)]

Numerator:

= √5 + √2 − √5 + √2 = 2√2

Denominator:

= 5 − 2 = 3

Answer: 2√2/3

6. Solving Equations Involving Surds

Surds can appear as coefficients of unknown variables. The first step is often to simplify the surds and collect the terms containing the unknown.

Worked Example 9

Solve x√8 = x√2 + 3.

Simplify √8:

2x√2 = x√2 + 3

Subtract x√2 from both sides:

x√2 = 3

x = 3/√2

Rationalise:

x = 3√2/2

Worked Example 10

Solve 2x√3 − √12 = x√3 + 4.

Since √12 = 2√3:

2x√3 − 2√3 = x√3 + 4

Collect the terms containing x:

x√3 = 4 + 2√3

Divide by √3:

x = (4 + 2√3)/√3

Separate the terms:

x = 4/√3 + 2

Rationalise:

x = 2 + 4√3/3

7. Equations That Require Squaring

Some equations contain an unknown inside a square root. Squaring can remove the root, but every resulting solution must be checked in the original equation.

Worked Example 11

Solve √(x + 1) = 3.

Square both sides:

x + 1 = 9

x = 8

Check in the original equation:

√(8 + 1) = √9 = 3

Therefore x = 8.

Worked Example 12

Solve √(x + 5) = x − 1.

Since the left side is non-negative, x − 1 must be non-negative. Thus x ≥ 1.

Square both sides:

x + 5 = (x − 1)²

x + 5 = x² − 2x + 1

x² − 3x − 4 = 0

(x − 4)(x + 1) = 0

So x = 4 or x = −1.

But x ≥ 1, so x = −1 is rejected.

Check x = 4:

√(4 + 5) = 3 and 4 − 1 = 3.

Therefore x = 4.

8. Surds in Exact Geometrical Values

Surds are useful in geometry because lengths and distances are often irrational. Leaving an answer in surd form gives an exact value rather than an approximation.

Worked Example 13

A square has side length 5 cm. Find the exact length of its diagonal.

Let the diagonal be d.

Using Pythagoras' theorem:

d² = 5² + 5²

d² = 25 + 25 = 50

d = √50

d = 5√2 cm

Worked Example 14

A rectangle has length 7 cm and width 4 cm. Find its diagonal exactly.

Let the diagonal be d.

d² = 7² + 4²

d² = 49 + 16 = 65

d = √65 cm

Since 65 has no square factor greater than 1, this is already simplified.

9. Surds and Mathematical Proof

Surds can be used in proofs to show that expressions are rational, irrational or equal to a required exact form.

Worked Example 15

Show that 1/(2 − √3) is equal to 2 + √3.

Multiply numerator and denominator by 2 + √3:

1/(2 − √3) = (2 + √3)/[(2 − √3)(2 + √3)]

= (2 + √3)/(4 − 3)

= 2 + √3

This proves the required result.

Worked Example 16

Show that (√3 + 1)² − (√3 − 1)² = 4√3.

Expand the first square:

(√3 + 1)² = 3 + 2√3 + 1 = 4 + 2√3

Expand the second square:

(√3 − 1)² = 3 − 2√3 + 1 = 4 − 2√3

Subtract:

= (4 + 2√3) − (4 − 2√3)

= 4√3

Hence the identity is proved.

10. Exact Values Versus Decimal Approximations

A surd is often preferred when an exact answer is required. A decimal approximation is useful when a measurement or practical estimate is needed.

Example:

Exact value: √2

Approximate value: √2 ≈ 1.4142

Exact value: 5√2 cm

Approximate value: 5√2 ≈ 7.071 cm

In an examination, follow the wording. If an exact answer is required, do not replace it with a rounded decimal.

11. Common Examination Mistakes

Mistake 1: Using the wrong conjugate

For 3 + √5, the conjugate is 3 − √5, not 3 + √5.

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Mistake 2: Forgetting to square both terms

(a + b)² = a² + 2ab + b².

It is not simply a² + b².

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Mistake 3: Losing a negative sign

When subtracting brackets, change the signs of every term inside the second bracket.

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Mistake 4: Accepting every squared-equation root

Squaring can introduce extraneous solutions. Always substitute into the original equation.

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Mistake 5: Giving an approximation when an exact answer is required

Keep the surd form unless the question requests a decimal answer.

12. Examination Practice

Attempt every question before opening the worked solutions. Show all necessary steps and give exact answers where appropriate.

Exercise A: Rationalisation

1. Rationalise 4/(√7 + √3).
2. Rationalise 5/(3 − √2).
3. Rationalise 6/(√11 − √5).
4. Express (2 + √5)/(2 − √5) in the form a + b√5.
5. Simplify 1/(3 − √2) + 1/(3 + √2).

Open worked solutions — Exercise A

1. Multiply by √7 − √3:

4(√7 − √3)/(7 − 3) = √7 − √3.

2. Multiply by 3 + √2:

5(3 + √2)/(9 − 2) = (15 + 5√2)/7.

3. Multiply by √11 + √5:

6(√11 + √5)/(11 − 5) = 3(√11 + √5)/4.

4. Multiply by 2 + √5:

(2 + √5)²/(4 − 5) = (9 + 4√5)/(−1).

Answer: −9 − 4√5.

5. Combine the fractions:

[(3 + √2) + (3 − √2)]/(9 − 2) = 6/7.

Exercise B: Algebraic Applications

1. Express (4 + √3)/(4 − √3) in the form a + b√3.
2. Simplify 1/(√7 − √3) − 1/(√7 + √3).
3. Solve x√12 = x√3 + 6.
4. Solve √(x + 6) = x, stating any restrictions.
5. Expand and simplify (2√3 − 1)(√3 + 4).

Open worked solutions — Exercise B

1. Multiply by 4 + √3:

(4 + √3)²/(16 − 3) = (19 + 8√3)/13.

2. Combine the fractions:

[(√7 + √3) − (√7 − √3)]/(7 − 3)

= 2√3/4 = √3/2.

3. Since √12 = 2√3:

2x√3 = x√3 + 6, so x√3 = 6.

x = 6/√3 = 2√3.

4. Since √(x + 6) = x, require x ≥ 0.

Square: x + 6 = x².

x² − x − 6 = 0.

(x − 3)(x + 2) = 0.

So x = 3 or x = −2. The restriction rejects −2.

Check x = 3: √9 = 3. Answer: x = 3.

5. Expand:

2√3 × √3 + 8√3 − √3 − 4

= 6 + 7√3 − 4 = 2 + 7√3.

Exercise C: Geometry and Proof

1. A square has a diagonal of 8√2 cm. Find its side length.
2. A right-angled triangle has perpendicular sides 3√2 cm and 2√5 cm. Find its hypotenuse in simplest surd form.
3. Show that 1/(√5 − 2) = √5 + 2.
4. Show that (√6 + √2)² = 8 + 4√3.
5. Explain why √12 + √27 can be expressed as a single surd term.

Open worked solutions — Exercise C

1. For a square, diagonal = side × √2.

Side = 8√2/√2 = 8 cm.

2. By Pythagoras:

h² = (3√2)² + (2√5)² = 18 + 20 = 38.

h = √38 cm.

3. Multiply by √5 + 2:

(√5 + 2)/[(√5 − 2)(√5 + 2)]

= (√5 + 2)/(5 − 4) = √5 + 2.

4. Expand:

6 + 2√12 + 2 = 8 + 4√3.

5. √12 = 2√3 and √27 = 3√3.

They are like surds, so their sum is 5√3.

13. Lesson Summary

• Use the conjugate to rationalise denominators containing two terms.
• Apply the difference of two squares carefully.
• Combine fractions using a common denominator.
• Simplify surd coefficients before solving equations.
• When squaring equations, check every solution in the original equation.
• Use surds to express exact geometrical lengths.
• Show each algebraic step clearly in proof questions.

Before Moving On

Make sure you can rationalise a denominator containing two surds, simplify a fraction into the form a + b√c, and solve an equation involving a square root.

These skills will help you handle more advanced algebraic questions with confidence.

MATHEMATICS HIPHOP

Making Mathematics Easier, One Concept at a Time.

Understand every step. Practise consistently. Build mathematical confidence.

Coming Next: Chapter 09 — Powers, Part 5: Applications and Consolidation of Indices and Surds.

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