ch.5.pdf - Chapter 5 Quadratic Expressions Chapter 5 Get Ready Chapter 5 Get Ready Question 1 Page 208 a 3 y has one term It is a monomial b 5 6a 3 has

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Unformatted text preview: Chapter 5 Quadratic Expressions Chapter 5 Get Ready Chapter 5 Get Ready Question 1 Page 208 a) −3 y has one term. It is a monomial. b) 5 + 6a 3 has two terms. It is a binomial. c) 6 x 2 + x − 1 has three terms. It is a trinomial. d) 8a 4 b 4 − 6a 3b 2 + 2ab 2 has three terms. It is a trinomial. e) 5d 3e − 7e has two terms. It is a binomial. f) 19m + 8n − 3 p has three terms. It is a trinomial. Chapter 5 Get Ready Question 2 Page 208 a) 9 + 5 y 5 − 4 y 2 + y has a degree of 5. b) 8a 3b 2 + 9a 2b − 6a 4b 2 has a degree of 6. c) 10 x 7 y 2 − 3x3 y 3 + 5 x 4 y 4 has a degree of 9. d) 6abc − 5a 2 bc 2 − 7 abc 2 has a degree of 5. MHR • Principles of Mathematics 10 Solutions 1 Chapter 5 Get Ready Question 3 Page 208 a) ( 5 x + 7 ) + ( 2 x − 11) = 5 x + 7 + 2 x − 11 = 7x − 4 b) ( 3b − 8 ) − ( 6b − 7 ) = ( 3b − 8 ) + ( −6b + 7 ) = 3b − 8 − 6b + 7 = −3b − 1 ( ) ( ) c) 5 x 2 + 6 x + 8 + 2 x 2 + 5 x − 9 = 5 x 2 + 6 x + 8 + 2 x 2 + 5 x − 9 = 7 x 2 + 11x − 1 ( ) ( ) ( ) ( d) 9 y 3 − 7 y 2 + 6 − 3 y 3 − 5 y 2 + 8 = 9 y 3 − 7 y 2 + 6 + −3 y 3 + 5 y 2 − 8 = 9 y3 − 7 y2 + 6 − 3y3 + 5 y2 − 8 = 6 y3 − 2 y2 − 2 ( ) ( ) e) 7 a 2 + 3a − 4 + 8a 2 − 2a − 15 = 7 a 2 + 3a − 4 + 8a 2 − 2a − 15 = 15a 2 + a − 19 ( ) ( ) ( ) ( f) 2c 2 − 3c + 1 − − c 2 − 3c − 5 = 2c 2 − 3c + 1 + c 2 + 3c + 5 = 2c − 3c + 1 + c + 3c + 5 2 2 = 3c 2 + 6 2 MHR • Principles of Mathematics 10 Solutions ) ) Chapter 5 Get Ready ( Question 4 ) ( Page 208 ) a) 7 x 2 + 3 xy − 2 y 2 + 8 x 2 − xy − y 2 = 7 x 2 + 3 xy − 2 y 2 + 8 x 2 − xy − y 2 = 15 x 2 + 2 xy − 3 y 2 ( ) ( ) ( ) ( b) 4 g 2 + gh − 7h 2 − g 2 − 2 gh + 3h 2 = 4 g 2 + gh − 7h 2 + − g 2 + 2 gh − 3h 2 ) = 4 g 2 + gh − 7h 2 − g 2 + 2 gh − 3h 2 = 3g 2 + 3gh − 10h 2 ( ) ( ) c) 5ab 2 + 7 a − b + 3ab 2 − 5a + 6b = 5ab 2 + 7 a − b + 3ab 2 − 5a + 6b = 8ab 2 + 2a + 5b ( ) ( ) ( ) ( d) 3cd 2 + 2c + 9d − 2cd 2 + 2c − d = 3cd 2 + 2c + 9d + −2cd 2 − 2c + d ) = 3cd + 2c + 9d − 2cd − 2c + d 2 2 = cd 2 + 10d e) ( 2 x + 8 ) − ( 6 x − 7 ) + ( 5 x − 1) = 2 x + 8 − 6 x + 7 + 5 x − 1 = x + 14 ( ) ( ) ( ) ( ) ( ) ( f) 5a 2 − b + 6b − 2a 2 − b2 + 7a 2 = 5a 2 − b + 6b − 2a 2 + −b2 − 7a 2 = 5a − b + 6b − 2a − b − 7a 2 2 2 ) 2 = −4a 2 + 5b − b2 MHR • Principles of Mathematics 10 Solutions 3 Chapter 5 Get Ready Question 5 Page 209 a) 3(x + 2) b) 4(x + 2) c) x(x + 3) d) 4x(x + 4) 4 MHR • Principles of Mathematics 10 Solutions Chapter 5 Get Ready Question 6 Page 209 b) −4 ( c + 9 ) = −4 ( c ) + ( −4 )( 9 ) a) 7m ( 3m + 8 ) = 7m ( 3m ) + 7m (8 ) = −4c − 36 = 21m + 56m 2 ( ) ( ) ( c) 5a 2 6a 2 − 8a = 5a 2 6a 2 + 5a 2 ( −8a ) = 30a 4 − 40a 3 Chapter 5 Get Ready ) ( ) d) 2 d 2 − 2d + 1 = 2 d 2 + 2 ( −2d ) + 2 (1) = 2d 2 − 4d + 2 Question 7 Page 209 a) V = ( 2 x )( 6 x )( 5 x − 1) = 12 x 2 ( 5 x − 1) = 12 x 2 ( 5 x ) + 12 x 2 ( −1) = 60 x 3 − 12 x 2 b) A = 2 ( 2 x )( 6 x ) + 2 ( 2 x )( 5 x − 1) + 2 ( 6 x )( 5 x − 1) = 24 x 2 + 4 x ( 5 x − 1) + 12 x ( 5 x − 1) = 24 x 2 + 4 x ( 5 x ) + 4 x ( −1) + 12 x ( 5 x ) + 12 x ( −1) = 24 x 2 + 20 x 2 − 4 x + 60 x 2 − 12 x = 104 x 2 − 16 x Chapter 5 Get Ready Question 8 Page 209 a) The factors of 10 are 1, 2, 5, and 10. b) The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. c) The factors of 16 are 1, 2, 4, 8, and 16. d) The factors of 32 are 1, 2, 4, 8, 16, and 32. Chapter 5 Get Ready Question 9 Page 209 a) 8 = 2 × 2 × 2 b) 14 = 2 × 7 c) 28 = 2 × 2 × 7 d) 30 = 2 × 3 × 5 MHR • Principles of Mathematics 10 Solutions 5 Chapter 5 Get Ready Question 10 Page 209 a) 6 = 2 × 3 9 = 3× 3 The GCF of 6 and 9 is 3. b) 25 = 5 × 5 15 = 3 × 5 The GCF of 15 and 25 is 5. c) 24 = 2 × 2 × 2 × 3 16 = 2 × 2 × 2 × 2 The GCF of 16 and 24 is 2 × 2 × 2 , or 8. d) 20 = 2 × 2 × 5 28 = 2 × 2 × 7 The GCF of 20 and 28 is 2 × 2 , or 4. e) 36 = 2 × 2 × 3 × 3 15 = 3 × 5 The GCF of 36 and 15 is 3. f) 32 = 2 × 2 × 2 × 2 × 2 40 = 2 × 2 × 2 × 5 The GCF of 32 and 40 is 2 × 2 × 2 , or 8. 6 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 1: Multiply Polynomials Chapter 5 Section 1 Question 1 Page 217 a) The model illustrates ( x + 1)( 2 x + 3) = 2 x 2 + 5 x + 3 b) The model illustrates ( x + 1)( x + 3) = x 2 + 4 x + 3 c) The model illustrates ( x + 2 )( x + 3) = x 2 + 4 x + 4 d) The model illustrates ( x + 3)( 2 x + 1) = 2 x 2 + 7 x + 3 MHR • Principles of Mathematics 10 Solutions 7 Chapter 5 Section 1 Question 2 Page 217 a) ( 2 x + 1)( x + 1) b) ( x + 4 )( x + 2 ) c) ( x + 1)( x + 5 ) 8 MHR • Principles of Mathematics 10 Solutions d) ( 2 x + 1)( 3 x + 2 ) Chapter 5 Section 1 Question 3 a) ( x + 3)( x + 5) = x ( x + 5) + 3 ( x + 5) Page 217 b) ( x + 3)( x + 4 ) = x ( x + 4 ) + 3 ( x + 4 ) = x + 5 x + 3x + 15 = x 2 + 4 x + 3x + 12 = x 2 + 8 x + 15 = x 2 + 7 x + 12 2 c) ( y + 2 )( y + 4 ) = y ( y + 4 ) + 2 ( y + 4 ) d) ( r + 4 )( r + 2 ) = r ( r + 2 ) + 4 ( r + 2 ) = y + 4y + 2y +8 = r 2 + 2r + 4r + 8 = y2 + 6 y + 8 = r 2 + 6r + 8 2 e) ( n + 7 )( n + 1) = n ( n + 1) + 7 ( n + 1) f) ( p + 9 )( p + 9 ) = p ( p + 9 ) + 9 ( p + 9 ) = n + n + 7n + 7 = p 2 + 9 p + 9 p + 81 = n 2 + 8n + 7 = p 2 + 18 p + 81 2 g) ( w + 7 )( w + 8 ) = w ( w + 8 ) + 7 ( w + 8 ) h) ( d + 3)( d + 11) = d ( d + 11) + 3 ( d + 11) = w2 + 8w + 7 w + 56 = d 2 + 11d + 3d + 33 = w2 + 15w + 56 = d 2 + 14d + 33 MHR • Principles of Mathematics 10 Solutions 9 Chapter 5 Section 1 Question 4 Page 217 b) ( y − 3)( y − 4 ) = y ( y − 4 ) − 3 ( y − 4 ) a) ( k − 3)( k − 5) = k ( k − 5) − 3 ( k − 5) = k − 5k − 3k + 15 = y 2 − 4 y − 3 y + 12 = k 2 − 8k + 15 = y 2 − 7 y + 12 2 d) ( q − 4 )( q − 2 ) = q ( q − 2 ) − 4 ( q − 2 ) c) ( x − 2 )( x − 4 ) = x ( x − 4 ) − 2 ( x − 4 ) = x2 − 4 x − 2 x + 8 = q2 − 2q − 4q + 8 = x2 − 6x + 8 = q 2 − 6q + 8 e) ( j − 7 )( j − 1) = j ( j − 1) − 7 ( j − 1) f) ( p − 9 )( p − 3) = p ( p − 3) − 9 ( p − 3) = j2 − j − 7 j + 7 = p 2 − 3 p − 9 p + 27 = j2 − 8 j + 7 = p 2 − 12 p + 27 g) ( z − 7 x )( z − 8 x ) = z ( z − 8 x ) − 7 x ( z − 8 x ) = z − 8 xz − 7 xz + 56 x 2 h) ( b − 3c )( b − 11c ) = b ( b − 11c ) − 3c ( b − 11c ) = b 2 − 11bc − 3bc + 33c 2 2 = z 2 − 15 xz + 56 x 2 Chapter 5 Section 1 = b 2 − 14bc + 33c 2 Question 5 Page 218 b) ( y + 3)( y − 4 ) = y ( y − 4 ) + 3 ( y − 4 ) a) ( x + 3)( x − 5) = x ( x − 5) + 3 ( x − 5) = x 2 − 5 x + 3 x − 15 = y 2 − 4 y + 3 y − 12 = x 2 − 2 x − 15 = y 2 − y − 12 c) ( c − 2 )( c + 4 ) = c ( c + 4 ) − 2 ( c + 4 ) d) ( w − 4 )( w + 2 ) = w ( w + 2 ) − 4 ( w + 2 ) = c + 4 c − 2c − 8 = w2 + 2 w − 4 w − 8 = c 2 + 2c − 8 = w2 − 2 w − 8 2 f) ( y − 9 )( y + 3) = y ( y + 3) − 9 ( y + 3) e) ( m + 7 )( m − 1) = m ( m − 1) + 7 ( m − 1) = m 2 − m + 7m − 7 = y 2 + 3 y − 9 y − 27 = m 2 + 6m − 7 = y 2 − 6 y − 27 g) ( x + 7 y )( x − 8 y ) = x ( x − 8 y ) + 7 y ( x − 8 y ) = x − 8 xy + 7 xy − 56 y 2 h) ( a + 6b )( a − 10b ) = a ( a − 10b ) + 6b ( a − 10b ) 2 = x 2 − xy − 56 y 2 10 MHR • Principles of Mathematics 10 Solutions = a 2 − 10ab + 6ab − 60b 2 = a 2 − 4ab − 60b 2 Chapter 5 Section 1 Question 6 Page 218 b) ( y − 3)( 5 y − 7 ) = 5 y 2 − 7 y − 15 y + 21 a) ( 2 x + 3)( x + 4 ) = 2 x 2 + 8 x + 3 x + 12 = 5 y 2 − 22 y + 21 = 2 x 2 + 11x + 12 c) ( 6c − 1) ( 3c + 5 ) = 18c 2 + 30c − 3c − 5 d) ( 7 w − 2 )( 2 w + 1) = 14 w2 + 7 w − 4 w − 2 = 18c 2 + 27c − 5 = 14 w2 + 3w − 2 f) ( 9 y − 2 )( 2 y + 2 ) = 18 y 2 + 18 y − 4 y − 4 e) ( 5m + 6 )( 5m − 6 ) = 25m 2 − 30m + 30m − 36 = 18 y 2 + 14 y − 4 = 25m 2 − 36 g) ( 7 d + 5c )( 8d − 6c ) = 56d 2 − 42cd + 40cd − 30c 2 = 56d 2 − 2cd − 30c 2 h) ( 6q + 5r )( 7 q − 12r ) = 42q 2 − 72qr + 35qr − 60r 2 = 42q 2 − 37 qr − 60r 2 Chapter 5 Section 1 Question 7 ( = 3( x a) 3 ( x − 5 )( x + 6 ) = 3 x 2 + 6 x − 5 x − 30 2 + x − 30 Page 218 ) ( = −2 ( x b) −2 ( x − 7 )( x − 9 ) = −2 x 2 − 9 x − 7 x + 63 ) = 3 x + 3 x − 90 − 6 y − 16 ) ( = 2(k ) = − y + 6 y + 16 − 8mn + 15n 2 2 ( = 2a (18a ) ) 2 f) 2a ( 3a + 4b )( 6a + 7b ) = 2a 18a 2 + 21ab + 24ab + 28b 2 2 ) + 10k + 21 = 2k + 20k + 42 = m − 8m n + 15mn 3 2 2 e) m ( m − 3n )( m − 5n ) = m m 2 − 5mn − 3mn + 15n 2 2 ) d) 2 ( k + 3)( k + 7 ) = 2 k 2 + 7 k + 3k + 21 2 ( = m(m ) 2 c) − ( y + 2 )( y − 8 ) = − y 2 − 8 y + 2 y − 16 2 − 16 x + 63 = −2 x + 32 x − 126 2 ( = −( y 2 ) + 45ab + 28b 2 ) ) = 36a 3 + 90a 2b + 56ab 2 MHR • Principles of Mathematics 10 Solutions 11 Chapter 5 Section 1 Question 8 Page 218 a) ( x + 4 )( x + 6 ) + ( x − 1)( x + 7 ) = x 2 + 6 x + 4 x + 24 + x 2 + 7 x − x − 7 = 2 x 2 + 16 x + 17 ( b) ( 2 x + 5 )( 3x − 7 ) + 2 ( 4 x + 9 )( 2 x − 11) = 6 x 2 − 14 x + 15 x − 35 + 2 8 x 2 − 44 x + 18 x − 99 ) = 6 x 2 − 14 x + 15 x − 35 + 16 x 2 − 88 x + 36 x − 198 = 22 x 2 − 51x − 233 ( = 3 ( 36 x ) ( − 18 x + 2 ) − 1(10 x c) 3 ( 6 x − 2 )( 6 x − 1) − ( 2 x − 3)( 5 x + 6 ) = 3 36 x 2 − 6 x − 12 x + 2 − 10 x 2 + 12 x − 15 x − 18 2 2 − 3 x − 18 ) ) = 108 x 2 − 54 x + 6 − 10 x 2 + 3x + 18 = 98 x 2 − 51x + 24 ( ) ( = −1( x − 5 x + 6 ) + 2 ( 3x + 17 x + 20 ) d) − ( x − 2 )( x − 3) + 2 ( 3x + 5)( x + 4 ) = − x 2 − 3x − 2 x + 6 + 2 3x 2 + 12 x + 5 x + 20 2 ) 2 = − x 2 + 5 x − 6 + 6 x 2 + 34 x + 40 = 5 x 2 + 39 x + 34 e) ( x + 4 ) − ( x − 4 ) = ( x + 4 )( x + 4 ) − ( x − 4 )( x − 4 ) 2 2 = x 2 + 4 x + 4 x + 16 − ( x 2 − 4 x − 4 x + 16 ) = x 2 + 8 x + 16 − 1( x 2 − 8 x + 16 ) = x 2 + 8 x + 16 − x 2 + 8 x − 16 = 16 x ( = −5 (15 x ) ( − 11x + 2 ) + 6 ( 30 x f) −5 ( 3x − 1)( 5 x − 2 ) + 6 ( 6 x + 3)( 5 x − 2 ) = −5 15 x 2 − 6 x − 5 x + 2 + 6 30 x 2 − 12 x + 15 x − 6 2 2 + 3x − 6 ) = −75 x 2 + 55 x − 10 + 180 x 2 + 18 x − 36 = 105 x 2 + 73x − 46 12 MHR • Principles of Mathematics 10 Solutions ) Chapter 5 Section 1 Question 9 a) h = −2 ( d − 3)( d − 15 ) ( = −2 ( d = −2 d 2 − 15d − 3d + 45 2 − 18d + 45 Page 218 ) ) = −2d + 36d − 90 2 b) h = −2 ( d − 3)( d − 15) h = −2d 2 + 36d − 90 = −2 (10 − 3)(10 − 15) = −2 (10 ) + 36 (10 ) − 90 = 70 = 70 2 Both forms of the equation predict a height of 70 m when d represents 10 m. Chapter 5 Section 1 Question 10 Page 218 a) A1 = x 2 b) A2 = ( x + 3)( x + 6 ) c) A2 = ( x + 3)( x + 6 ) = x 2 + 6 x + 3 x + 18 = x 2 + 9 x + 18 d) A2 − A1 = x 2 + 9 x + 18 − x 2 = 9 x + 18 e) A2 − A1 = 9 x + 18 = 9 (12 ) + 18 = 126 If x represents 12 m, the increase in area is 126 m2. MHR • Principles of Mathematics 10 Solutions 13 Chapter 5 Section 1 Question 11 Page 218 Question 12 Page 218 a) i) A = x ( x + 10 ) = x 2 + 10 x ii) A = x ( 2 x ) = 2 x2 iii) A = ( x + 5 )( x + 6 ) = x 2 + 6 x + 5 x + 30 = x 2 + 11x + 30 b) Chapter 5 Section 1 a) The x-intercepts of y = ( x + 3)( x − 1) are –3 and 1. b) y = ( x + 3)( x − 1) = x 2 − x + 3x − 3 = x2 + 2 x − 3 c) 14 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 1 Question 13 Page 219 Question 14 Page 219 a) b) V = w ( w + 2 )( 2 ) c) V = w ( w + 2 )( 2 ) = 2w ( w + 2 ) = 2 w2 + 4 w Chapter 5 Section 1 a) b) SA1 = 6 ( x )( x ) = 6 x2 c) SA2 = 6 ( x + y ) 2 d) SA2 − SA1 = 6 ( x + y ) − 6 x 2 2 = 6 ( x + y )( x + y ) − 6 x 2 ( = 6( x ) = 6 x 2 + xy + xy + y 2 − 6 x 2 2 ) + 2 xy + y 2 − 6 x 2 = 6 x + 12 xy + 6 y − 6 x 2 2 2 = 12 xy + 6 y 2 MHR • Principles of Mathematics 10 Solutions 15 e) V2 − V1 = ( x + y ) − x3 3 = ( x + y )( x + y )( x + y ) − x3 ( = ( x + y)( x ) = ( x + y ) x 2 + xy + xy + y 2 − x3 2 ) + 2 xy + y 2 − x3 = x + 2 x y + xy + x y + 2 xy 2 + y 3 − x3 3 2 2 2 = 3 x 2 y + 3xy 2 + y 3 Chapter 5 Section 1 Question 15 a) d = 3000 ( v + 0.3)(1.0 − v ) ( = 3000 ( −v = 3000 1.0v − v 2 + 0.3 − 0.3v 2 + 0.7v + 0.3 ) Page 219 ) = −3000v + 2100v + 900 2 b) d = −3000v 2 + 2100v + 900 = −3000 ( 0.2 ) + 2100 ( 0.2 ) + 900 2 = 1200 At 0.2 m/s, she can swim 1200 m before running out of air. c) She can swim a maximum of 1267.5 m at 0.35 m/s before running out of air. 16 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 1 Question 16 Page 219 a) Method 1: 4 ( x + 6 ) + 5 ( x ) = 4 x + 24 + 5 x = 9 x + 24 Method 2: ( x + 4 )( x + 6 ) − x ( x + 1) = x 2 + 6 x + 4 x + 24 − x 2 − x = 9 x + 24 b) Method 1: x ( x + 4 ) + 2 ( 3) = x 2 + 4 x + 6 Method 2: x ( x + 7 ) − 3( x − 2 ) = x 2 + 7 x − 3x + 6 = x2 + 4 x + 6 Chapter 5 Section 1 a) Question 17 Page 219 n = 500 − 100 p n − 500 = −100 p n − 500 =p −100 500 n − + =p 100 100 n =p 5− 100 b) R = np n ⎞ ⎛ = n⎜5 − ⎟ ⎝ 100 ⎠ n2 = 5n − 100 c) MHR • Principles of Mathematics 10 Solutions 17 Chapter 5 Section 1 Question 18 Page 219 Question 19 Page 219 Question 20 Page 219 s = n2 − n + 1 Chapter 5 Section 1 Answers will vary. Chapter 5 Section 1 The number n must be 18 more than a multiple of 24. The smallest possible value for n is 24 + 18 = 42. If 42 is divided by 8, the remainder is 2. Answer C 18 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 2 Special Products Chapter 5 Section 2 Question 1 Page 225 a) b) c) d) MHR • Principles of Mathematics 10 Solutions 19 Chapter 5 Section 2 Question 2 Page 225 Use the appropriate pattern for squaring a binomial. a) ( x + 5) = ( x ) + 2 ( x )( 5) + ( 5) 2 2 b) ( y + 4 ) = ( y ) + 2 ( y )( 4 ) + ( 4 ) 2 2 2 = y 2 + 8 y + 16 = x 2 + 10 x + 25 c) ( w + 6 ) = ( w ) + 2 ( w )( 6 ) + ( 6 ) 2 2 d) ( k + 7 ) = ( k ) + 2 ( k )( 7 ) + ( 7 ) 2 2 2 = w2 + 12 w + 36 2 f) ( c + 10 ) = ( c ) + 2 ( c )(10 ) + (10 ) 2 2 = m 2 + 22m + 121 2 2 h) ( x + 20 ) = ( x ) + 2 ( x )( 20 ) + ( 20 ) 2 2 2 = g 2 + 18 g + 81 Chapter 5 Section 2 2 = c 2 + 20c + 100 g) ( g + 9 ) = ( g ) + 2 ( g )( 9 ) + ( 9 ) 2 2 = k 2 + 14k + 49 e) ( m + 11) = ( m ) + 2 ( m )(11) + (11) 2 2 2 = x 2 + 40 x + 400 Question 3 Page 225 Use the appropriate pattern for squaring a binomial. 2 2 2 2 2 2 a) ( x − 5) = ( x ) − 2 ( x )( 5) + ( 5) b) ( z − 3) = ( z ) − 2 ( z )( 3) + ( 3) = x 2 − 10 x + 25 = z2 − 6z + 9 c) ( x − 9 ) = ( x ) − 2 ( x )( 9 ) + ( 9 ) 2 2 d) ( c − 1) = ( c ) − 2 ( c )(1) + (1) 2 2 = x 2 − 18 x + 81 2 f) ( b − 100 ) = ( b ) − 2 ( b )(100 ) + (100 ) 2 2 2 g) ( n − 2 ) = ( n ) − 2 ( n )( 2 ) + ( 2 ) 2 h) ( m − 6 ) = ( m ) − 2 ( m )( 6 ) + ( 6 ) 2 2 = n 2 − 4n + 4 Chapter 5 Section 2 2 = b2 − 200b + 10 000 = v 2 − 24v + 144 2 2 = c 2 − 2c + 1 e) ( v − 12 ) = ( v ) − 2 ( v )(12 ) + (12 ) 2 2 2 2 = m 2 − 12m + 36 Question 4 Page 225 Use the appropriate pattern for squaring a binomial. a) ( x + 3 y ) = ( x ) + 2 ( x )( 3 y ) + ( 3 y ) 2 2 b) ( 2 x − y ) = ( 2 x ) − 2 ( 2 x )( y ) + ( y ) 2 2 2 = x 2 + 6 xy + 9 y 2 c) ( 5c + 2d ) = ( 5c ) + 2 ( 5c )( 2d ) + ( 2d ) 2 2 = 4 x 2 − 4 xy + y 2 d) ( 3a − 4b ) = ( 3a ) − 2 ( 3a )( 4b ) + ( 4b ) 2 2 = 25c 2 + 20cd + 4d 2 e) ( 9k + 2m ) = ( 9k ) + 2 ( 9k )( 2m ) + ( 2m ) 2 2 2 2 = 9a 2 − 24ab + 16b2 2 f) ( 4u − 5v ) = ( 4u ) − 2 ( 4u )( 5v ) + ( 5v ) = 81k 2 + 36km + 4m 2 20 MHR • Principles of Mathematics 10 Solutions 2 2 = 16u 2 − 40uv + 25v 2 2 2 Chapter 5 Section 2 Question 5 Page 225 Use the appropriate pattern for the product of a sum and a difference. 2 2 2 2 a) ( v + 1)( v − 1) = ( v ) − (1) b) ( a − 1) ( a + 1) = ( a ) − (1) = v2 − 1 = a2 − 1 c) ( y + 5) ( y − 5) = ( y ) − ( 5) 2 d) ( x − 7 ) ( x + 7 ) = ( x ) − ( 7 ) 2 2 = y 2 − 25 e) ( e − 9 ) ( e + 9 ) = ( e ) − ( 9 ) 2 2 = x 2 − 49 f) ( z + 6 ) ( z − 6 ) = ( z ) − ( 6 ) 2 2 = e 2 81 2 = z 2 − 36 g) ( x + 12 )( x − 12 ) = ( x ) − (12 ) 2 h) ( y − 3) ( y + 3) = ( y ) − ( 3) 2 2 = y2 − 9 = x 2 − 144 Chapter 5 Section 2 2 Question 6 Page 225 Use the appropriate pattern for the product of a sum and a difference. 2 2 2 2 a) ( w − v )( w + v ) = ( w ) − ( v ) b) ( 3m − n ) ( 3m + n ) = ( 3m ) − ( n ) = w2 − v 2 = 9m 2 − n 2 c) ( y + 6 x ) ( y − 6 x ) = ( y ) − ( 6 x ) 2 d) ( 3x + 4 y ) ( 3x − 4 y ) = ( 3x ) − ( 4 y ) 2 2 = y 2 − 36 x 2 e) ( 7 g − 3h )( 7 g + 3h ) = ( 7 g ) − ( 3h ) 2 = 49 g 2 − 9h 2 2 = 9 x 2 − 16 y 2 2 f) ( 9 x − 8 y )( 9 x + 8 y ) = ( 9 x ) − (8 y ) 2 2 = 81x 2 − 64 y 2 MHR • Principles of Mathematics 10 Solutions 21 Chapter 5 Section 2 a) Question 7 ( x + 4 )( x − 4 ) = ( x ) − ( 4 ) 2 Page 225 2 = x 2 − 16 ( x + 4 )( x − 4 ) = ( 2 + 4 )( 2 − 4 ) = −12 x 2 − 16 = 22 − 16 = −12 b) ( x − 8) 2 = ( x ) − 2 ( x )(8 ) + (8 ) 2 2 = x 2 − 16 x + 64 ( x − 8) 2 = ( 2 − 8) 2 = 36 x 2 − 16 x + 64 = 22 − 16 ( 2 ) + 64 = 36 c) ( x + 8) 2 = ( x ) + 2 ( x )(8 ) + ( 8 ) 2 2 = x 2 + 16 x + 64 ( x + 8) 2 = ( 2 + 8) 2 = 100 x 2 + 16 x + 64 = 2 2 + 16 ( 2 ) + 64 = 10 0 22 MHR • Principles of Mathematics 10 Solutions d) ( x − 10 )( x + 10 ) = ( x ) − (10 ) 2 2 = x 2 − 100 ( x − 10 )( x + 10 ) = ( 2 − 10 )( 2 + 10 ) = −96 x 2 − 100 = 22 − 100 = −96 e) ( x + 11)( x − 11) = ( x ) − (11) 2 2 = x 2 − 121 ( x + 11)( x − 11) = ( 2 + 11)( 2 − 11) = −117 x 2 − 121 = 22 − 121 = −117 f) ( x + 12 ) 2 = ( x ) + 2 ( x )(12 ) + (12 ) 2 2 = x 2 + 24 x + 144 ( x + 12 ) 2 = ( 2 + 12 ) 2 = 196 x 2 + 24 x + 144 = 22 + 24 ( 2 ) + 144 = 196 g) ( x − 7) 2 = ( x ) − 2 ( x )( 7 ) + ( 7 ) 2 2 = x 2 − 14 x + 49 ( x − 7) 2 = (2 − 7) 2 = 25 x 2 − 14 x + 49 = 22 − 14 ( 2 ) + 49 = 25 MHR • Principles of Mathematics 10 Solutions 23 h) ( x − 30 )( x + 30 ) = ( x ) − ( 30 ) 2 2 = x 2 − 900 ( x − 30 )( x + 30 ) = ( 2 − 30 )( 2 + 30 ) = −896 x 2 − 900 = 22 − 900 = −896 Chapter 5 Section 2 A = π (r + k ) Question 8 Page 226 Question 9 Page 226 2 2 2 = π ⎡( r ) + 2 ( r )( k ) + ( k ) ⎤ ⎣ ⎦ = π ( r 2 + 2rk + k 2 ) = π r 2 + 2π rk + π k 2 Chapter 5 Section 2 a) b) A = ( x + 5 )( x + 5 ) = x 2 + 10 x + 25 c) Increase in area = x 2 + 10 x + 25 − x 2 = 10 x + 25 24 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 2 Question 10 Page 226 a) The vertex is (–2, 0). b) y = ( x + 2 ) 2 = ( x ) + 2 ( x )( 2 ) + ( 2 ) 2 2 = x2 + 4 x + 4 c) L.S. = y R.S. = x 2 + 4 x + 4 =0 = ( −2 ) + 4 ( −2 ) + 4 2 =0 L.S. = R.S. The point (–2, 0) satisfies the equation y = x 2 + 4 x + 4 . Chapter 5 Section 2 Question 11 Page 226 a) A = ( 3x + 2 y )( 3x − 2 y ) = ( 3x ) − ( 2 y ) 2 2 = 9 x2 − 4 y2 b) Change in area = ( 3x + 2 y )( 3x − 2 y ) − ( 3x ) 2 = 9 x2 − 4 y2 − 9 x2 = −4 y 2 c) A = 9 x2 − 4 y2 = 9 (8 ) − 4 ( 5) 2 2 = 476 Change in area = −4 y 2 = −4 ( 5 ) 2 = −100 The area of the rectangle is 476 cm2. The change in area is 100 cm2 less. MHR • Principles of Mathematics 10 Solutions 25 Chapter 5 Section 2 Question 12 Page 226 Question 13 Page 226 Method 1: A = ( x + 2 )( x − 2 ) + 2 ( 4 ) = ( x ) − (2) + 8 2 2 = x2 − 4 + 8 = x2 + 4 Method 2: A = ( x − 2) + 4x 2 = ( x ) − 2 ( x )( 2 ) + ( 2 ) + 4 x 2 2 = x2 − 4x + 4 + 4x = x2 + 4 The expressions are equivalent. Chapter 5 Section 2 a) b) 2 2 Exposed area = ( 2 x + 5) − ( 2 x + 2 ) 2 2 2 2 = ( 2 x ) + 2 ( 2 x )( 5) + ( 5) − ⎡ ( 2 x ) + 2 ( 2 x )( 2 ) + ( 2 ) ⎤ ⎣ ⎦ 2 2 = 4 x + 20 x + 25 − 4 x − 8 x − 4 = 12 x + 21 c) Exposed area = ( 2 x + 2 ) − ( 2 x − 1) 2 2 2 2 2 2 = ( 2 x ) + 2 ( 2 x )( 2 ) + ( 2 ) − ⎡ ( 2 x ) − 2 ( 2 x )(1) + (1) ⎤ ⎣ ⎦ 2 2 = 4 x + 8x + 4 − 4 x + 4 x − 1 = 12 x + 3 26 MHR • Principles of Mathematics 10 Solutions Chapter 5 Section 2 Question 14 a) 31 × 29 = ( 30 + 1)( 30 − 1) Page 226 b) 59 × 61 = ( 60 − 1)( 60 + 1) = 30 − 1 = 602 − 12 = 900 − 1 = 899 = 3600 − 1 = 3599 2 2 c) 99 × 101 = (100 − 1)(100 + 1) d) 71 × 69 = ( 70 + 1)( 70 − 1) = 100 − 1 = 702 − 12 = 10 000 − 1 = 9999 = 4900 − 1 = 4899 2 2 Chapter 5 Section 2 Question 15 Page 226 32 × 28 = ( 30 + 2 )( 30 − 2 ) = 302 − 22 = 900 − 4 = 896 a) 76 × 84 = ( 80 − 4 )( 80 + 4 ) = 80 − 4 2 b) 35 × 25 = ( 30 + 5 )( 30 − 5 ) = 302 − 52 2 = 6400 − 16 = 6384 = 900 − 25 = 875 c) 104 × 96 = (100 + 4 )(100 − 4 ) d) 77 × 83 = ( 80 − 3)( 80 + 3) = 1002 − 42 = 802 − 32 = 10 000 − 16 = 9984 = 6400 − 9 = 6391 Chapter 5 Section 2 Question 16 Page 227 a) b) h = −5 ( t − 3) + 10 2 MHR • Principles of Mathematics 10 Solutions 27 c) h = −5 ( t − 3) + 10 2 2 2 = −5 ⎡ ( t ) + 2 ( t )( 3) + ( 3) ⎤ + 10 ⎣ ⎦ = −5 ( t 2 − 6t + 9 ) + 10 = −5t 2 + 30t − 45 + 10 = 5t 2 + 30t − 35 Chapter 5 Section 2 Question 17 Page 227 a) R = ( 3264 + x )( 2448 + x ) = 7 990 272 + 3264 x + 2448 x + x 2 = 7 990 272 + 5712 x + x 2 b) R = 7 990 272 + 5712 x + x 2 = 7 990 272 + 5712 (1000 ) + (1000 ) 2 = 14 702 272 The resolution of the new CCD is 14.7 megapixels. Chapter 5 Section 2 Question 18 Page 227 Solutions for Achievement Checks are shown in the Teacher’s Resource. Chapter 5 Section 2 Question 19 a) ( x − 2 ) = ( x − 2 ) ( x − 2 ) 4 2 ( Page 227 2 )( = x2 − 4 x + 4 x2 − 4 x + 4 ) = x − 4 x + 4 x − 4 x + 16 x 2 − 16 x + 4 x 2 − 16 x + 16 4 3 2 3 = x 4 − 8 x 3 + 24 x 2 − 32 x + 16 ( = ( 2 x + 3) ( 4 x b) ( 2 x + 3)( x − 5 )( 4 x + 7 ) = ( 2 x + 3) 4 x 2 + 7 x − 20 x − 35 2 − 13 x − 35 ) ) = 8 x − 26 x − 70...
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  • Summer '18
  • jane
  • Math, Binomial, Harshad number, Orders of magnitude, 170, perfect square trinomial

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