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Applications of Expansion and Factorisation SLIDESHOW 17 MATHEMATICS MR SASAKI ROOM 307 Objectives β’Use the difference of two squares to make certain numerical calculations β’Be able to prove certain simple numerical facts The Difference of Two Squares What is the difference of two squares? The law states that if an expression π + π is multiplied by its conjugate, we get the difference of π2 and π 2 . π + π π β π β‘ π2 β π 2 for β π, π β β. This rule has many uses to help make calculations easier. The Difference of Two Squares π + π π β π β‘ π2 β π 2 for β π, π β β. Letβs use this identity to help us make some numerical calculations. Example Calcuate 51 × 49. Consider the above for π = 50, π = 1. 51 × 49 = 50 + 1 50 β 1 = 502 β 12 = 2500 β 1 = 2499 The Difference of Two Squares We also need to consider the opposite principle. π2 β π 2 β‘ π + π π β π for β π, π β β. Example Calcuate 452 β 52 . Consider the above for π = 45, π = 5. 452 β 52 = 45 + 5 45 β 5 = 50 × 40 = 2000 Answers 6399 396 2491 2451 8019 3536 999900 8096 48.91 7 πππ 9 600 1800 6600 900000 3800000 β9000 Numerical Proofs Before we look at proofs, we should recall some vocabulary. Odd - A number in the form 2π β 1 where π β β€+ . (1, 3, 5, 7, β¦) Even - A number in the form 2π where π β β€+ . (2, 4, 6, 8, β¦) Consecutive Integers A pair of integers π, π + 1 where π β β€. (for example 15 and 16.) Proofs (Format) When you prove something, there should be three steps: Step 1 - A statement about what you have to prove. (eg: Let an even number be a number be in the form 2π where π β β€+ .) Step 2 - Perform the necessary calculation. Write what you need to in the correct form. Step 3 - State that as itβs now in the correct form, the proof for the original expression is now complete. Numerical Proofs Letβs try a simple proof and divide the steps up. Example Prove that an odd number squared is odd. Step 1 Let an odd number π = 2π β 1 where π β β€+ . Step 2 If π = 2π β 1, π2 = 2π β 1 2 = 4π2 β 4π + 1 2 = 2(2π β 2π) + 1 Step 3 As π2 = 2 2π2 β 2π + 1 is in the form 2π΄ + 1 where π΄ β β€+ , π2 must be odd. β΄ If π is odd, π2 must be odd. Numerical Proofs Letβs try another. Example If two integers are odd, their sum is even. Step 1 Consider two integers π₯, π¦ where π₯ = 2π β 1, π¦ = 2π β 1 and π, π β β€+ . Step 2 The sum of π₯ and π¦, π₯ + π¦ = 2π β 1 + (2π β 1) = 2π + 2π β 2 = 2(π + π β 1) Step 3 As π₯ + π¦ is in the form 2π΄ where π΄ β β€+ , π₯ + π¦ is even. β΄ If π₯ and π¦ are odd, π₯ + π¦ must be even. Answers β Easy (Top) Let an odd number be in the form 2π β 1 where π β β€+ . 7 = 2 β 4 β 1 where 4 β β€+ . As 7 can be written in the form 2π β 1 where π β β€+ , 7 is odd. Let an even number be in the form 2π where π β β€+ . 12 = 2 β 6 where 6 β β€+ . As 12 can be written in the form 2π where π β β€+ , 12 is even. Let an even number π = 2π where π β β€+ . If π = 2π, π2 = 2π 2 = 4π2 = 2 2π2 . As π2 is in the form 2π΄ where π΄ β β€+ , π2 is even. β΄ An even number squared is even. Answers β Easy (Bottom) Let an even number π = 2π where π β β€+ . If π = 2π, π π = π β β β€+ as π β β€+ .β΄ Half of an even number 2 2 is an integer. Consider two integers π₯, π¦ where π₯ = 2π, π¦ = 2π and π, π β β€+ . The sum of π₯ and π¦, π₯ + π¦ = 2π + 2π = 2 π + π . As π₯ + π¦ is in the form 2π΄ where π΄ β β€+ , π₯ + π¦ is even.β΄ If two integers are even, their sum is even. Let π be an odd number where π = 2π β 1, π β β€+ . As π2 = 2π β 1 2, π2 = 4π2 β 4π + 1 = 2 2π2 β 2π + 1. As π2 is in the form 2π΄ + 1 where π΄ β β€+ , π2 is odd. β΄ The positive root of a square odd number is odd. Answers β Hard (Top) Let a pair of consecutive odd numbers π = 2π β 1, π = 2π + 1, π β β€+. ππ + 1 = 2π β 1 2π + 1 + 1 = 4π2. As ππ + 1 is in the form 4π΄ where π΄ β β€+, the product of two consecutive odd numbers plus 1 is a multiple of 4. Consider two integers π₯, π¦ where π₯ = 2π β 1, π¦ = 2π and π, π β β€+. The sum of π₯ and π¦, π₯ + π¦ = 2π β 1 + 2π = 2 π + π β 1. As π₯ + π¦ is in the form 2π΄ β 1where π΄ β β€+, π₯ + π¦ is odd. β΄ The sum of an odd and even number is odd. Consider two integers π₯, π¦ where π₯ = 2π β 1, π¦ = 2π β 1 and π, π β β€+. The product of π₯ and π¦, π₯π¦ = (2π β Answers β Hard (Bottom) Consider two integers π₯, π¦ where π₯ = 2π β 1, π¦ = 2π and π, π β β€+ . The product of π₯ and π¦, π₯π¦ = 2π 2π β 1 = 2(2ππ β π). As π₯π¦ is in the form 2π΄ where π΄ β β€+ , the expression is even. β΄ The product of an odd and even number is even. Let an odd number π be in the form π = 2π β 1 where π β β€+ . By substitution, 3π + 7 = 3 2π β 1 + 7 = 6π + 4 = 2 3π + 2 . As 3π + 7 is in the form 2π΄ where π΄ β β€+ , the expression is even. β΄ If π is odd, 3π + 7 is even.