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Quadratic Equations Solving Quadratic Equations

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7/9/2013 Quadratic Equations 2 Solving Quadratic Equations Standard Form a x 2 + b x + c = 0 with a ≠ 0 Solve to Find: Solutions and Solution Set What is a solution ? What is a solution set ? How many solutions ?

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7/9/2013 Quadratic Equations 3 Solving Quadratic Equations Standard Form a x 2 + b x + c = 0 with a ≠ 0 Solve by: Square Root Property Completing the Square Quadratic Formula Factoring

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7/9/2013 Quadratic Equations 4 Solving Quadratic Equations Solutions A solution for the equation a x 2 + b x + c = 0 is a value of variable x satisfying the equation NOTE: For now we assume all values are real numbers The solution set for the equation is the set of all of its solutions

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7/9/2013 Quadratic Equations 5 Solving Quadratic Equations Examples 1. x 2 + 2x – 3 = 0 2. 16x 2 – 8x + 1 = 0 3. x 2 + 2x + 5 = 0 Solutions: –3, 1 Solution set: { –3, 1 } Solution: 1 4 Solutions: none Solution set: { } OR Solution set: { } 1 4

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7/9/2013 Quadratic Equations 6 The Square Root Property Reduced Standard Form - Type 1 a x 2 = c, for a ≠ 0 Rewrite: Square Root Property If x 2 = k then OR c a x2x2 = = k x2x2 = x │ k = x k = x –

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7/9/2013 Quadratic Equations 7 The Square Root Property Square Root Property If x 2 = k then OR x2x2 = x │ k = x k = x – Note: This does NOT say that This DOES say that IF x 2 = 4 then OR The Principal Root The Negative Root = 2 x = 4 2 4 = 4 – x == –2–2

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7/9/2013 Quadratic Equations 8 Square Root Property Roof repairs on a 100-story building A tar bucket slips and falls to the sidewalk If each story is 12.96 feet, how long does it take the bucket to hit the sidewalk? Distance of fall: (12.96 ft/story)(100 stories) = 1296 ft For a fall of s ft in t seconds we have s = 16t 2 OR t 2 = s/16

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7/9/2013 Quadratic Equations 9 Square Root Property Roof repairs on a 100-story building Solving for t We choose the positive value of t Thus s = 16t 2 OR t 2 = s/16 OR = t s 4 – = t 4 s WHY ? 4 36 = 9 = seconds = t 4 s = 4 1296

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7/9/2013 Quadratic Equations 10 Special Forms Perfect-Square Trinomials Always the square of a binomial A 2 + 2AB + B 2 = (A + B) 2 A 2 – 2AB + B 2 = (A – B) 2 … OR (B – A) 2

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7/9/2013 Quadratic Equations 11 Special Forms Examples x 2 + 6x + 9 4y 2 – 20y + 25 9y 2 + 12xy + 4x 2 x 2 y 2 – 6xyz + 9z 2 = (x + 3) 2 = (2y – 5) 2 = (3y + 2x) 2 = (xy – 3z) 2

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7/9/2013 Quadratic Equations 12 Completing the Square Convert to perfect-square trinomial Solve using square root property Example 1: 2x 2 – 3x + 5 = x 2 – 9x + 12 x 2 + 6x = 7 To make the left side we need to find a such that (x + a ) 2 = x 2 + 2 a x + a 2 Subtract x 2 and 5, on each side, and add 9x = x 2 + 6x + a 2 a perfect square

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7/9/2013 Quadratic Equations 13 Completing the Square Convert to perfect-square trinomial Example 1: 2x 2 – 3x + 5 = x 2 – 9x + 12 x 2 + 6x = 7 (x + a ) 2 = x 2 + 2 a x + a 2 = x 2 + 6x + a 2 So 2 a x = 6x giving a = 3 and a 2 = 9 Adding 9 to each side of x 2 + 6x = 7 x 2 + 6x + 9 = 7 + 9 = 16 (x + 3) 2 = 16x + 3 16= + = + 4 x = –3 + 4 Solution set: { –7, 1 }

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7/9/2013 Quadratic Equations 14 Completing the Square Convert to perfect-square trinomial Example 2: 5y 2 – 87y + 89 = 25 + 15y – 4y 2 9y 2 – 102y = –64 For the left need to find a such that (3y + a ) 2 = 9y 2 + 6 a y + a 2 = 9y 2 – 102y + a 2 Add 4y 2 to each side, subtract 89 and 15y to be a perfect square we

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7/9/2013 Quadratic Equations 15 Completing the Square Example 2: 5y 2 – 87y + 89 = 25 + 15y – 4y 2 9y 2 – 102y = –64 (3y + a ) 2 = 9y 2 + 6 a y + a 2 = 9y 2 – 102y + a 2 So 6 a y = –102y giving a = –17 and a 2 = 289 Adding 289 to each side 9y 2 – 102y + 289 = –64 + 289 Thus (3y – 17) 2 = 225

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7/9/2013 Quadratic Equations 16 Completing the Square Example 2: 5y 2 – 87y + 89 = 25 + 15y – 4y 2 9y 2 – 102y = –64 (3y – 17) 2 = 225 3y – 17 225= + 15= + 3y = 17 15 + y = + 17 3 15 3 = 32 3 2 3 Solution Set: { 32 3 2 3, }

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7/9/2013 Quadratic Equations 17 The Quadratic Formula I Complete the Square on the Standard Form a x 2 + b x + c = 0, for a ≠ 0 Rewrite: move c, divide by a c a b a – x2x2 x + = Now complete the square on the left side Want a constant k such that b a x2x2 x + + k2+ k2 is a perfect square – that is b a x2x2 x + + k2+ k2 = x + k ( ) 2

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7/9/2013 Quadratic Equations 18 The Quadratic Formula II Complete the Square on the Standard Form c a b a – x2x2 x + = b a x2x2 x + + k2+ k2 = x + k ( ) 2 = x 2 + 2xk + k 2 Thus 2k = b a … and = b 2a2a ( ) 2 k2k2 b a x2x2 x + c a – = + b 2a2a ( ) 2 + b 2a2a ( ) 2 x + b 2a2a ( ) 2 c a – = + b 4a4a 2 2

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7/9/2013 Quadratic Equations 19 The Quadratic Formula III Complete the Square on the Standard Form x + ( b 2a2a ) 2 = c a – + b 2 4a4a 2 + b 2 4a4a 2 = – 4 ac 4a4a 2 = 4a4a 2 b 2 – x + b 2a2a = 4a4a 2 b 2 – + x = b 2a2a 2a2a b 2 – + x = 2a2a b b 2 – + Square Root Property Quadratic Formula

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7/9/2013 Quadratic Equations 20 Quadratic Formula Examples Example 1 Solve: 7x 2 + 9x + 29 = 27 First put into standard form: 7x 2 + 9x + 2 = 0 x = 2a2a b 2 – 4 ac ± –b–b Apply Quadratic Formula Thus a = 7, b = 9, c = 2 WHY ? x = ± –9–9 9 2 – 4 7 2 (( ) ) 2 7 ( )

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7/9/2013 Quadratic Equations 21 Quadratic Formula Examples Example 1 Solve: 7x 2 + 9x + 29 = 27 x = ± –9–9 9 2 – 4 7 2 (( ) ) 2 7 ( ) = ± –9–9 81 – 56 14 = ± –9–9 2525 = ± –9–9 5 Question:Are there always two solutions ? Solution set: { – 1, } 7 2 – = –14 14 (continued) – 4– 4 14

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7/9/2013 Quadratic Equations 22 Quadratic Formula Examples Example 2 Solve: 4x 2 – 12x + 29 = 20 First put into standard form: 4x 2 – 12x + 9 = 0 WHY ? Thus a = 4, b = –12, c = 9 x = 2a2a b 2 – 4 ac ± –b–b Apply Quadratic Formula x = 2 4 ( ) –12 2 – 4 4 9 ± 12 (( ) ) ()

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7/9/2013 Quadratic Equations 23 Quadratic Formula Examples Example 2 Solve: 4x 2 – 12x + 29 = 20 Solution set: x = 2 4 ( ) –12 2 – 4 4 9 ± 12 (( ) ) () (continued) x = 8 144 – 144 ± 12 = 8 x = 2 3 x { 2 3 } Question:Are there always two solutions ?

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7/9/2013 Quadratic Equations 24 The Quadratic Formula The Discriminant For a x 2 + b x + c = 0 we found that The expression b 2 – 4 ac is called the discriminant It determines the number and type of solutions x = 2a2a b 2 – 4 ac ± –b–b

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7/9/2013 Quadratic Equations 25 The Quadratic Formula The Discriminant Determines the number and type of solutions If b 2 – 4 ac = 0 have one real solution If b 2 – 4 ac > 0 have two real solutions If b 2 – 4 ac < 0 have two complex solutions

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7/9/2013 Quadratic Equations 26 Factoring Reduced Standard Form – Type 2 a x 2 + b x = 0, for a ≠ 0 Factor x and use zero-product property x( a x + b ) = 0 Zero-product Property: if and only if p = 0 or q = 0... or both For any real numbers p and q, pq = 0

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7/9/2013 Quadratic Equations 27 Factoring Reduced Standard Form – Type 2 a x 2 + b x = 0, for a ≠ 0 Example: Solve 3x 2 – 5x = 0 x(3x – 5) = 0 From the zero-product property x = 0 or 3x – 5 = 0 or both Solution Set: { 0, 5 3 }

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7/9/2013 Quadratic Equations 28 Special Forms Conjugate Binomials Pairs of binomials of form (A + B) and (A – B) are called conjugate binomials Product of conjugate pair is always a difference of squares (A + B)(A – B) = A 2 – B 2

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7/9/2013 Quadratic Equations 29 Special Forms Conjugate Binomials Examples x 2 – 4 9 – 4x 2 4x 2 – 9y 2 = (x + 2)(x – 2) = (3 – 2x)(3 + 2x) = (2x – 3y)(2x + 3y)

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7/9/2013 Quadratic Equations 30 Not-So-Special Forms What if the expression is not a special form? We look for factors with integer coefficients In general these have form: x 2 + ( a + b )x + ab = (x + a )(x + b ) We look for factors (x + a ) and (x + b ) that fit this form

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7/9/2013 Quadratic Equations 31 Not-So-Special Trinomials What if the expression is not a special form? Factoring Examples: x 2 + 7x + 12 (x + 3)(x + 4) = y 2 + 13y + 40(y + 5)(y + 8) = (2x + 3)(2x + 5) = 4x 2 + 16x + 15 Question: Is there a systematic way to do this ?

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7/9/2013 Quadratic Equations 32 Not-So-Special Trinomials Factoring Trinomials with negative constant term In general these have form: x 2 + ( a – b )x – ab = (x + a )(x – b ) We look for factors that fit this form

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7/9/2013 Quadratic Equations 33 Not-So-Special Trinomials Examples x 2 – x – 12 = y 2 + 3y – 40 = 4x 2 – 4x – 15 = Note the sign of last term: always negative Note the sign of the middle term: positive or negative (x + 3)(x – 4) (y – 5)(y + 8) (2x + 3)(2x – 5) WHY ?

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7/9/2013 Quadratic Equations 34 Factor x 2 – 5x – 24 x 2 + ( a + b )x + ab = (x + a )(x + b ) Trinomials with two negative terms Not-So-Special Trinomials Terms a, b of opposite sign Sign of the larger term Find a and b such that Possible factors of –24 : 1, –24 2, –12 3, –8 4, –6 Since 3 – 8 = –5, then a = 3 and b = –8

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7/9/2013 Quadratic Equations 35 Factor x 2 – 5x – 24 Trinomials with two negative terms Not-So-Special Trinomials Since 3 – 8 = –5, then a = 3 and b = –8 This gives us: NOTE: There is no solution to talk about here … since we are changing the form of an expression, NOT solving an equation x 2 – 5x – 24 = (x + 3)(x – 8)

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7/9/2013 Quadratic Equations 36 Factor x 2 + 5x – 24 x 2 + ( a + b )x + ab = (x + a )(x + b ) Trinomials with one negative term – the constant Not-So-Special Trinomials Terms a, b of opposite sign Sign of the larger term Find a and b such that Possible factors of –24 : –1, 24 –2, 12 –3, 8 –4, 6 Since 3 – 8 = 5, then a = 8 and b = –3

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7/9/2013 Quadratic Equations 37 Factor x 2 + 5x – 24 Trinomials with one negative term – the constant Not-So-Special Trinomials Since 3 – 8 = 5, then a = 8 and b = –3 This gives us: x 2 + 5x – 24 = (x + 8)(x – 3) NOTE: This is not a solution, since we are changing the form of an expression, NOT solving an equation

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7/9/2013 Quadratic Equations 38 Solve x 2 – 8x + 12 = 0 by factoring x 2 + ( a + b )x + ab = (x + a )(x + b ) Trinomials with one negative term – the middle term Not-So-Special Trinomials Terms a, b of same sign Sign of both terms Find a and b such that Possible factors of 12 : –1, –12 –2, –6 –3, –4 Since –2 – 6 = –8, then a = –2 and b = –6

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7/9/2013 Quadratic Equations 39 Solve x 2 – 8x + 12 = 0 by factoring Trinomials with one negative term – the middle term Not-So-Special Trinomials Since –2 – 6 = –8, then a = –2 and b = –6 x 2 – 8x + 12 = (x – 2)(x – 6) = 0 This gives us: NOTE: This is a solution, since we are solving an equation So x – 2 = 0 or x – 6 = 0 Solution Set: 2, 6 { }

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7/9/2013 Quadratic Equations 40 Solve x 2 + 11x + 28 = 0 by factoring x 2 + ( a + b )x + ab = (x + a )(x + b ) Trinomials with no negative terms Not-So-Special Trinomials Terms a, b of same sign Sign of both terms Find a and b such that Possible factors of 28 : 1, 28 2, 14 4, 7 Since 4 + 7 = 11, then a = 4 and b = 7

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7/9/2013 Quadratic Equations 41 Solve x 2 + 11x + 28 = 0 by factoring Trinomials with no negative terms Not-So-Special Trinomials Since 4 + 7 = 11, then a = 4 and b = 7 x 2 + 11x + 28 = (x + 4)(x + 7) = 0 This gives us: So x + 4 = 0 or x + 7 = 0 Solution Set: – 4, –7 { } NOTE: This is a solution, since we are solving an equation

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7/9/2013 Quadratic Equations 42 Expanding a Room A 9 by 12 room is to be expanded by the same amount x in length and width to double the area – what is x ? 9 + x 12 ft 9 ftA 1 = 108 A 4 = x 2 A 3 = 12x A 2 = 9x x ft 12 + x Original area: A 1 = 208 ft 2 New area: A = 2A 2 = 216 ft 2 Two methods: A = A 1 + A 2 + A 3 + A 4 A = (9 + x)(12 + x) = 216

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7/9/2013 Quadratic Equations 43 Expanding a Room Two methods: A = A 1 + A 2 + A 3 + A 4 A = (9 + x)(12 + x) = 216 OR Either way: x 2 + 21x + 108 = 216 Sox 2 + 21x – 108 = 0 ≈ 4.3 sq. ft. x = 2 1 ( ) (( ) ) () ± 21 2 – 4 1 –108 21 – x = 2 ± 876 21 –

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7/9/2013 Quadratic Equations 44 Think about it !

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7/9/2013 Quadratic Equations 45 The Quadratic Formula 1 Complete the Square on the Standard Form a x 2 + b x + c = 0, for a ≠ 0 Rewrite: move c, divide by a c a b a – x2x2 x + = Now complete the square on the left side Want a constant k such that b a x2x2 x + + k2+ k2 is a perfect square – that is b a x2x2 x + + k2+ k2 + = x2x2 x + b 2a2a ( ) 2 b 2a2a ( ) 2

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7/9/2013 Quadratic Equations 46 The Quadratic Formula 2 Complete the Square on the Standard Form b a x2x2 x + + k2+ k2 + = x2x2 x + b 2a2a ( ) 2 b 2a2a ( ) 2 c a b a – x2x2 x + = Thus k = b 2a2a …and = b 2a2a ( ) 2 k2k2 So … + k2k2 b a x2x2 x + c a – = + k2k2 + x2x2 x + b 2a2a ( ) 2 b 2a2a ( ) 2 = b 2a2a ( ) 2 c a – + x + ( b 2a2a ) 2 = c a – + b 2 4a4a 2

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7/9/2013 Quadratic Equations 47 The Quadratic Formula 3 Complete the Square on the Standard Form x + ( b 2a2a ) 2 = c a – + b 2 4a4a 2 + b 2 4a4a 2 = – 4 ac 4a4a 2 = 4a4a 2 b 2 – x + b 2a2a = 4a4a 2 b 2 – + x = b 2a2a 2a2a b 2 – + x = 2a2a b b 2 – + Square Root Property Quadratic Formula

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7/9/2013 Quadratic Equations 48 Want k so that (x + k) 2 = x 2 + 2kx + k 2 a b = x 2 + x + k 2 2k = a b, k = 2a2a b, k 2 = 2a2a b ( ) 2

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