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Math Lesson 7.5.1 - Multiplying Brackets containing Surds

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Welcome to our Math lesson on Multiplying Brackets containing Surds, this is the first lesson of our suite of math lessons covering the topic of Rationalising the Denominator, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Multiplying Brackets containing Surds

In tutorials 6.3 and 6.4 we discussed the eight special algebraic identities that help us factorise or simplify various expressions. All those identities are also useful when dealing with surds, especially the first three of them, i.e.

(a + b)2 = a2 + 2ab + b2
(a-b)2 = a2 -2ab + b2

and

(a-b)(a + b) = a2 -b2

The third special identity is particularly useful as it allows us getting rid of surds immediately, without the need for further operations. For example, in the expression

(5-√2)(5 + √2)

we can get rid of roots quickly, i.e.

(5 - √2)(5 + √2)
= 52 - (√2)2
= 25 - 2
= 23

However, the first two special algebraic identities are also useful as if we have let's say the sum or difference of two surds raised in the second power, the result will contain a single surd. For example,

(√3 + √2)2
= (√3)2 + 2 ∙ √3 ∙ √2 + (√2)2
= 3 + 2√(3 ∙ 2) + 2
= 5 + 2√6

The rest of algebraic identities are also useful in some situations. For example, the expression

11√2 + 9√3

is the simplified form of

(√2 + √3)3

because given the fourth algebraic identity (a + b)3 = a3 + 3a2b + 3ab2 + b3, we can write

(√2 + √3)3
= (√2)3 + 3 ∙ (√2)2 ∙ √3 + 3 ∙ √2 ∙ (√3)2 + (√3)3
= (√2)2 ∙ √2 + 3 ∙ (√2)2 ∙ √3 + 3 ∙ √2 ∙ (√3)2 + (√3)2 ∙ √3
= 2√2 + 3 ∙ 2 ∙ √3 + 3 ∙ √2 ∙ 3 + 3√3
= 2√2 + 6√3 + 9√2 + 3√3
= 11√2 + 9√3

Example 1

Write the following expressions in the simplest form.

  1. (3 - √7)(3 + √7)
  2. (√6 - 1)2 + (√2 + √3)2

Solution 1

  1. Applying the formula of the third special algebraic identity, yields:
    (3 - √7)(3 + √7)
    = 32 - (√7)2
    = 9 - 7
    = 2
  2. Combining the formulae of the first two special algebraic identities yields
    (√6 - 1)2 + (√2 + √3)2
    = (√6)2 - 2 ∙ √6 ∙ 1 + 12 + (√2)2 + 2 ∙ √2 ∙ √3 + (√3)2
    = 6 - 2√6 + 1 + 2 + 2√6 + 3
    = 12

More Rationalising the Denominator Lessons and Learning Resources

Powers and Roots Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
7.5Rationalising the Denominator
Lesson IDMath Lesson TitleLessonVideo
Lesson
7.5.1Multiplying Brackets containing Surds
7.5.2Rationalising the Denominator

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