Expanding Brackets with Surds and Fractions Slideshow 9, Mr Richard Sasaki, Mathematics

Objectives Be able to expand brackets with surds Expanding brackets with surds on the outside Calculate with surds in fractions

Expanding Brackets (Linear) Let’s think back to algebra. When we expand brackets, we multiply terms on the inside by the one on the outside. 3𝑥 2𝑥−𝑦 = 6 𝑥 2 −3𝑥𝑦 The same principles apply with surds. 2( 2 −3)= 2 2 −6 In this case, the expression cannot be simplified. But sometimes we are able to.

Expanding Brackets (Linear) Let’s try an example where we can simplify. Example Expand and simplify 4(2 3 + 12 ). 4 2 3 + 12 = 8 3 +4 12 =8 3 +4∙2∙ 3 =8 3 +8 3 =16 3 Note: We could simplify initially but then there would be no need to expand.

32 2 20 11 8 3 +40 2 28+14 3 4 6 +2 2 5 6 −18 2 6+2 3 3− 6 6 5 −5 10− 5 10+ 15 7+2 7 + 14 −14−4 7 11−2 11 240−45 2 36 70 +18 10 +12 2

Surds in Fractions We had a look at some surd fractions in the form 𝑎 𝑏 𝑐 𝑑 where 𝑎, 𝑏, 𝑐, 𝑑∈ℤ (𝑐, 𝑑≠0). Let’s review. Example Simplify 1 2 3 . 3 2 3 ∙ 3 = 3 6 1 2 3 = Remember, a fraction should have an integer as its denominator.

Surds in Fractions Questions with different denominators require a different thought process. We need to expand brackets. Example Simplify 4 3 +2 3 − 2 7 −5 4 . 4 3 +2 3 − 2 7 −5 4 = 4(4 3 +2) 3∙4 − 3(2 7 −5) 4∙3 = 16 3 +8 12 − 6 7 −15 12 = 16 3 +8−6 7 +15 12 = 16 3 −6 7 +23 12

5 +2 3 +7 4 2 7 −6 3 +21 6 9 2 −4 7 −4 5 +27 12 4 6 +9 6 23 5 +3 6 7 3 −5 7 +73 35 35 3 −8 6 28 95 3 +42 2 6

Roots in Denominators Calculating with roots in denominators requires us to expand brackets where roots are on the outside. Example Simplify 2 +1 3 − 3 −1 2 . 2 +1 3 − 3 −1 2 = 3 2 +1 3 ∙ 3 − 2 3 −1 2 ∙ 2 = 6 + 3 3 − 6 − 2 2 = 2 6 + 3 6 − 3 6 − 2 6 = 2 6 +2 3 6 − 3 6 −3 2 6 = 2 3 +3 2 − 6 6

11 5 −9 6 27 11 44 2 5 +5 3 5 5 14 +2 6 4 17 3 −69 5 3 651 2 −82 10 35

4 15 +4 3 3 13 2 −4 72 3 −6 2 +12 3 4 3 +9 2 −6 18 6 5 +63 2 −160 30 10 5 −9 7 +254 15