Section 1.5 Equations

Transcription

1 Section 1.5 Equations Linear and Rational Equations EXAMPLES: 1. Solve the equation 7x 4 = x+. 7x 4 = x+ 7x 4+4 = x++4 7x = x+1 7x x = x+1 x 4x = 1 4x 4 = 1 4 x = or, in short, 7x 4 = x+ 7x x = +4 4x = 1 x = 1 4 =. Solve the equation 5(x 4)+ = (x+7). 5(x 4)+ = (x+7) 5x+0+ = x+14 5x+ = x+11 5x+ x = x+11 x 7x+ = 11 7x+ = 11 7x = 11 7x 7 = 11 7 x = 11 7 or, in short, 5(x 4)+ = (x+7) 5x+0+ = x+14 5x x = x = 11 x =

2 . Solve the equation y +5 y 4 In short, = y y +5 y = y ( y +5 1 y ) ( y +7 = y +5 1 y 4 = 1 y +7 6(y +5) (y ) = 4(y +7)+1 6y +0 y +6 = 4y ++1 y +6 = 4y +40 y +6 4y = 4y +40 4y y +6 = 40 y +6 6 = 40 6 y = 4 y( 1) = 4( 1) y = 4 y +5 y = y ( y +5 1 y ) ( y +7 = 1 4 6(y +5) (y ) = 4(y +7)+1 6y +0 y +6 = 4y = 4y 6y +y y = 4 ) ) Solve the equation 15 y = 1 4y +.

3 4. Solve the equation 15 y = 1 4y y = 1 4y + 4y 15 y = 4y 4y 15 y ( 1 4y + ) 1 = 4y +4y 4y 4 15 = 1+y 60 = 1+y 60 1 = 1+y 1 9 = y 9 = y y = 9 or, in short, 15 y = 1 4y + 4y 15 y = 4y 60 = 1+y 9 = y y = 9 ( 1 4y + ) 5. Solve the equation y y +5 = 5 y (y +5) 4(y +5) y y +5 = 5 y y +5 y = 4(y +5) ( 5 y ) 4 y y +5 = 4(y +5) (y +5) y y = 4 ( 5)+(y +5) 5 4y = 0+5y +5 4y = 5+5y 4y 5y = 5+5y 5y y = 5 ( 1)( y) = ( 1)5 y = 5

4 In short, 6. Solve the equation (x+4)(x+1) 4(y +5) y y +5 = 5 y y +5 y = 4(y +5) ( 5 y ) 4 4y = 4 ( 5)+(y +5) 5 4y = 0+5y = 5y 4y y = 5 11 x +5x+4 x+4 = 1 x+1. (x+4)(x+1) 11 x +5x+4 x+4 = 1 x+1 11 (x+4)(x+1) x+4 = 1 x+1 ( 11 (x+4)(x+1) ) x+4 = (x+4)(x+1) 1 x+1 11 (x+4)(x+1) (x+4)(x+1) x+4 = (x+4)(x+1) 1 x+1 11 (x+1) = (x+4) 1 11 x = x+4 x = x+4 x = x+4 x = x 4 x x = x 4 x 4x = 4 4x 4 = 4 4 x = 1 4

5 In short, (x+4)(x+1) 11 x +5x+4 x+4 = 1 x+1 11 (x+4)(x+1) x+4 = 1 x+1 ( 11 (x+4)(x+1) ) x+4 = (x+4)(x+1) 11 (x+1) = (x+4) 1 11 x = x+4 x x = x = 4 x = 1 1 x+1 7. Solve for M the equation F = G mm r. ( ) Gm F = M = r The solution is M = r F Gm. ( r Gm ) ( r F = Gm )( Gm r ) M = r F Gm = M. The surface area A of the closed rectangular box can be calculated from the length l, the width w, and the height h according to the formula A = lw +wh+lh Solve for w in terms of the other variables in this equation. 5

6 . The surface area A of the closed rectangular box can be calculated from the length l, the width w, and the height h according to the formula A = lw +wh+lh Solve for w in terms of the other variables in this equation. The solution is w = A lh l+h. A = lw +wh+lh A = (l+h)w+lh A lh = (l+h)w A lh l+h = w Quadratic Equations EXAMPLES: 1. Solve each equation: (a) x = 0 (b) x = 1 (c) x = 4 (d) x = 5 (e) (x 4) = 7 6

7 1. Solve each equation: (a) x = 0 (b) x = 1 (c) x = 4 (d) x = 5 (e) (x 4) = 7 Solution: (a) We have x = 0. (b) We have (see the Appendix) x = ±1. (c) We have (see the Appendix) x = ± (which is ± 4). (d) We have (see the Appendix) x = ± 5. (e) We have (x 4) = 7 The solutions are x = 4 7 and x = Solve the equation x +5x = 4. The solutions are x = and x =.. Solve each equation: x 4 = ± 7 [ x 4+4 = ± ] 7+4 x = 4± 7 x +5x = 4 x +5x 4 = 0 (x )(x+) = 0 x = 0 or x+ = 0 x = or x = (a) x x+1 = 0 (b) x 1x+7 = 0 7

8 . Solve each equation: (a) x x+1 = 0 (b) x 1x+7 = 0 Solution: (a) We have x x+1= 0 x x+1 1= 0 1 x x= 1 x x 4= 1 x x 4+4 = 1+4 (x 4) = x 4= ± x 4+4= ± +4 x= 4± In short, x x+1 = 0 x x 4 = 1 x x 4+4 = 1+4 (x 4) = x 4 = ± x = 4± The solutions are x = 4 and x = 4+.

9 (b) We have x 1x+7= 0 x 1x+7 7= 0 7 x 1x= 7 x 1x = 7 x 1x = 7 x 4x= 7 x x = 7 In short, x x + = 7 + { 7 (x ) = 5 x = ± 5 x += ± + 5 x= ± 7 +4 = = = = 7+1 x 1x+7 = 0 x 1x = 7 } = 5 x 4x = 7 x x + = 7 + The solutions are x = 5 and x = + (x ) = 5 5 x = ± 5. x = ± 5 9

10 Proof: We have ax +bx+c= 0 ax +bx+c c= 0 c ax +bx= c ax +bx = c a a ax a + bx a = c a x +x x + b a x= c a x +x b a = c a ( ) b b a + = c ( ) b a a + a ( x+ a) b { c = a + b (a) = c a + b a = c a + b 4a = 4ac + }= b 4ac+b 4a 4a 4a x+ b 4ac+b 4ac+b {± a = = ± 4a 4a x+ b a b a = ± b 4ac b a a = ± } b 4ac b 4ac = ± 4 a a x= b a ± b 4ac a = b± b 4ac a 10

11 In short, ax +bx+c = 0 ax +bx = c x +x x + b a x = c a ( ) b b a + = c ( ) b a a + a ( x+ b ) = 4ac+b a 4a x+ b a = ± b 4ac a x = b± b 4ac a EXAMPLES: 1. Solve the equation x 4x 5 = 0. Solution: We first rewrite the equation as x +( 4)x+( 5) = 0. Here a =, b = 4, and c = 5. Therefore by the quadratic formula, x = b± b 4ac a = ( 4)± ( 4) 4 ( 5) = 4± = 4± 76 = 4± 4 19 = 4± 4 19 = ± 19 = (± 19) = ± 19 In short, x = ( 4)± ( 4) 4 ( 5) = 4± 76 = 4± 4 19 = ± 19 = ± 19 11

12 . Solve the equation x = 4. Solution: We first rewrite the equation as 1 x +0 x+( 4) = 0 Here a = 1, b = 0, and c = 4. Therefore by the quadratic formula, x = b± b 4ac a = 0± ( 4) 1 = ± 0+16 = ± 16 = ±4 = ±. Find all solutions of each equation. (a) x 5x 1 = 0 (b) 4x +1x+9 = 0 (c) x +x+ = 0 1

13 . Find all solutions of each equation. (a) x 5x 1 = 0 (b) 4x +1x+9 = 0 (c) x +x+ = 0 Solution: (a) We first rewrite the equation as x +( 5)x+( 1) = 0. Here a =, b = 5, and c = 1. Therefore by the quadratic formula, x = b± b 4ac a = ( 5)± ( 5) 4 ( 1) = 5± = 5± 7 6 (b) In this quadratic equation a = 4, b = 1, and c = 9. Therefore by the quadratic formula, x = b± b 4ac a = 1± = 1± = 1± 0 = 1±0 = 1 = (c) In this quadratic equation a = 1, b =, and c =. Therefore by the quadratic formula, x = b± b 4ac a = ± 4 1 = ± 4 = ± 4 Since the square of any real number is nonnegative, 4 is undefined in the real number system. The equation has no real solution. EXAMPLES: 1. Use the discriminant to determine how many real solutions each equation has. (a) x +4x 1 = 0 (b) 4x 1x+9 = 0 (c) 1 x x+4 = 0 1

14 1. Use the discriminant to determine how many real solutions each equation has. (a) x +4x 1 = 0 (b) 4x 1x+9 = 0 (c) 1 x x+4 = 0 Solution: (a) We first rewrite the equation as 1 x + 4x + ( 1) = 0. Here a = 1, b = 4, and c = 1. Therefore the discriminant is D = b 4ac = ( 1) = 16+4 = 0 > 0 so the equation has two distinct real solutions. (b) We first rewrite the equation as 4x +( 1)x+9 = 0. Here a = 4, b = 1, and c = 9. Therefore the discriminant is D = b 4ac = ( 1) = = 0 so the equation has exactly one real solution. (c) We first rewrite the equation as 1 x + ( )x + 4 = 0. Here a = 1, b =, and c = 4. Therefore the discriminant is D = b 4ac = ( ) = { so the equation has no real solution. = = = 1 16 = 1 16 } = 4 < 0. An object thrown or fired straight upward at an initial speed of v 0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h = 16t +v 0 t Suppose that a bullet is shot straight upward with an initial speed of 00 ft/s. Its path is shown in the Figure below. (a) When does the bullet fall back to ground level? (b) When does it reach a height of 6400 ft? (c) When does it reach a height of mi? (d) How high is the highest point the bullet reaches? 14

15 . An object thrown or fired straight upward at an initial speed of v 0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h = 16t +v 0 t Suppose that a bullet is shot straight upward with an initial speed of 00 ft/s. Its path is shown in the Figure below. (a) When does the bullet fall back to ground level? Solution: Since the initial speed in this case is v 0 = 00 ft/s, the formula is h = 16t +00t Ground level corresponds to h = 0, so we must solve the equation 0 = 16t +00t 0 = 16t(t 50) Thus, t = 0 or t = 50. This means the bullet starts (t = 0) at ground level and returns to ground level after 50 s. 15

16 (b) When does it reach a height of 6400 ft? Solution: Setting h = 6400 in h = 16t +00t gives 16t 00t+6400 = 0 t 50t+400 = 0 (t 10)(t 40) = = 16t +00t t = 10 or t = 40 The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to earth). (c) When does it reach a height of mi? Solution: Two miles is 50 = 10,560 ft. Setting h = 10,560 in h = 16t +00t gives The discriminant of this equation is 16t 00t+10,560 = 0 10,560 = 16t +00t t 50t+660 = 0 D = b 4ac = ( 50) = 140 which is negative. Thus, the equation has no real solution. The bullet never reaches a height of mi. 16

17 (d) How high is the highest point the bullet reaches? Solution: Each height the bullet reaches is attained twice, once on its ascent and once on its descent. The only exception is the highest point of its path, which is reached only once. This means that for the highest value of h, the following equation has only one solution for t: 16t 00t+h = 0 h = 16t +00t This in turn means that the discriminant D of the equation is 0, and so The maximum height reached is 10,000 ft. D = b 4ac = ( 00) 4 16 h = 0 640,000 64h = 0 h = 10,000 EXAMPLES: 1. Solve the equation x = x. Other Types of Equations Solution 1: We have x = x Solution : If x 0, then x = x x x = 0 x x = x x x(x 1) = 0 x = 1 x(x 1)(x+1) = 0 x = ±1 x = 0 or x 1 = 0 or x+1 = 0 x = 0 or x = 1 or x = 1 Note that x = 0 is also a solution of the equation. This gives the same result.. Solve the following equations: (a) x 6 = 16x (b) x 7 = 7x 4 17

18 . Solve the following equations: (a) x 6 = 16x (b) x 7 = 7x 4 Solution 1(a): We have Solution (a): If x 0, then x 6 = 16x x 6 = 16x x 6 16x = 0 x = 16x x x (x 4 16) = 0 x 4 = 16 ( ) 4 x (x ) 4 = 0 x4 = 4 16 x (x 4)(x +4) = 0 x = x (x )(x+)(x +4) = 0 x = ± Since x +4 0, we have Note that x = 0 is also a solution of the equation. This x = 0 or x = 0 or x+ = 0 gives the same result. x = 0 or x = or x = Solution 1(b): We have x 6 Solution (b): If x 0, then x 7 = 7x 4 x 7 = 7x 4 x 7 7x 4 = 0 x = 7x4 4 x 4 x 4 (x 7) = 0 x = 7 x 4 (x ) = 0 x = 7 x 4 (x )(x +x+9) = 0 x = Since the discriminant of x +x+9 is D = = 7 < 0, it follows that x + x+9 0. Therefore x 4 = 0 or x = 0 x = 0 or x =. Solve the equation 0a 1a 45a+7 = 0. 0a 1a 45a+7 = 0 4a (5a ) 9(5a ) = 0 (5a )(4a 9) = 0 ) (5a ) ((a) = 0 (5a )(a )(a+) = 0 5a = 0 or a = 0 or a+ = 0 5a = or a = or a = x 7 Note that x = 0 is also a solution of the equation. This gives the same result. a = 5 or a = or a = 1

19 4. Solve the equation x + 5 x+ =. x + 5 x+ = ( x + 5 ) x(x+) = x(x+) x+ x x(x+)+ 5 x+ x(x+) = x +4x To solve x x = 0 we can either factor (x+)+5x = x +4x x+6+5x = x +4x x x = 0 (x )(x+1) = 0 0 = x +4x x 6 5x 0 = x 4x 6 0 = x x x = 0 or x+1 = 0 x = or x = 1 or use the quadratic formula with a = 1,b =, and c = : x = b± b 4ac a so = ( )± ( ) 4 1 ( ) 1 = ± 4+1 = ± 16 = ±4 x = +4 = 6 4 = or x = = = 1 The values x = and x = 1 are only potential solutions. We must check them to see if they satisfy the original equation. Check: x = x + 5 x+ = + 5 +? = 1+1 = TRUE Check: x = 1 x + 5 x+ = ? = +5 = TRUE We see that both x = and x = 1 are the solutions of the equation x + 5 x+ =. 19

20 5. Solve the equation x = 1 x. x = 1 x x 1 = x (x 1) = ( x) (x) x 1+1 = x 4x 4x+1 = x 4x 4x+1 +x = 0 4x x 1 = 0 To solve 4x x 1 = 0 we can either factor 4x x 1 = 0 (4x+1)(x 1) = 0 4x+1 = 0 or x 1 = 0 x = 1 or x = 1 4 or use the quadratic formula with a = 4,b =, and c = 1: x = b± b 4ac a so = ( )± ( ) 4 4 ( 1) 4 x = 5 = = 1 4 = ± 9+16 or x = +5 = = 1 = ± 5 = ±5 The values x = 1 and x = 1 are only potential solutions. We must check them to see if they 4 satisfy the original equation. Check: x = 1 4 ( 1 4 x = 1 x )? = 1? = 1 1 1? = ( = 1 TRUE ) Check: x = 1 x = 1 x 1 =? 1 1 =? 1 1 = 1 1 FALSE We see that x = 1 4 is a solution but x = 1 is not. So, the only solution is x = Solve the equation 6+x x+7 =. 0

21 6. Solve the equation 6+x x+7 =. Solution 1: We have 6+x x+7 = 6+x = + x+7 ( 6+x) = ( + x+7) 6+x = ( ) +( ) x+7+( x+7) 6+x = 4 4 x+7+x+7 6+x 4 x 7 = 4 x+7 x 5 = 4 x+7 (x 5) = ( 4 x+7) x x 5+5 = 16(x+7) x 10x+5 = 16x+11 x 10x+5 16x 11 = 0 x 6x 7 = 0 To solve x 6x 7 = 0 we can either factor x 6x 7 = 0 (x 9)(x+) = 0 x 9 = 0 or x+ = 0 x = 9 or x = or use the quadratic formula with a = 1,b = 6, and c = 7: x = b± b 4ac a so = ( 6)± ( 6) 4 1 ( 7) 1 = 6± = 6± 104 = 6± x = 6+ = 5 6 = 9 or x = = 6 = The values x = 9 and x = are only potential solutions. We must check them to see if they satisfy the original equation. Check: x = 9 6+x x+7 = ? = 64 6? = Check: x = 6+x x+7 = 6+ ( ) +7? = 0 4? = 6 = FALSE 0 = TRUE We see that x = is a solution but x = 9 is not. So, the only solution is x =. 1

22 Solution : We have 6+x x+7 = ( 6+x x+7) = ( ) ( 6+x) 6+x x+7+( x+7) = 4 and the same result follows. 6+x (6+x)(x+7)+x+7 = 4 7. Solve the equation x 4 x + = 0. Solution: Setting W = x, we get (6+x)(x+7) = 4 6 x x 7 }{{}}{{} 6x+4+x +14x=x +0x+4 x 9 x +0x+4 = x 9 x +0x+4 = x+9 ( x +0x+4) = (x+9) 4(x +0x+4) = (x) + x 9+9 x +0x+16 = 9x +54x+1 x 4 x + = 0 (x ) x + = 0 W W + = 0 0 = 9x +54x+1 x 0x 16 0 = x 6x 7 To solve this equation, we use the quadratic formula with a = 1,b =, and c = : so W = b± b 4ac a = ± It follows that there are four solutions: 4+ 4 = ( )± ( ) 4 = ± 64 = ± 16 = ± 16 = ±4 = ± 4 = 4± x = 4± = x = ± 4± Solve the equation x 1/ +x 1/6 = 0.

23 . Solve the equation x 1/ +x 1/6 = 0. Solution: Setting W = x 1/6, we get x 1/ +x 1/6 = 0 (x 1/6 ) +x 1/6 = 0 W +W = 0 (W 1)(W +) = 0 W 1 = 0 or W + = 0 W = 1 x 1/6 = 1 W = x 1/6 = (x 1/6 ) 6 = 1 6 (x 1/6 ) 6 = ( ) 6 x = 1 x = 64 The values x = 1 and x = 64 are only potential solutions. We must check them to see if they satisfy the original equation. Check: x = 1 Check: x = 64 x 1/ +x 1/6 = 0 1 1/ +1 1/6? = = 0 TRUE x 1/ +x 1/6 = / +64 1/6? = 0 4+ = 0 FALSE We see x = 1 is a solution but x = 64 is not. So, the only solution is x = 1.

24 Absolute Value Equations EXAMPLES: 1. Solve the equation x 5 =. Solution: By the definition of absolute value, x 5 = is equivalent to The solutions are x = 4 and x = 1.. Solve the equation 5 4x + = 5. x 5 = or x 5 = x = x = x = 4 x = 1 5 4x + = 5 5 4x = 5 }{{} 50 4x = 50 5 = 10 By the definition of absolute value, 4x = 10 is equivalent to 4x = 10 or 4x = 10 4x = 10 }{{} x = 4 = The solutions are x = and x =. 4x = 10 }{{} 1 x = 1 4 =. Solve the equation x+7 = x 4. Solution: By the definition of absolute value, x+7 = x 4 is equivalent to x+7 = x 4 or x+7 = (x 4) The solutions are x = 11 x x = 4 7 x = 11 x = 11 = 11 and x =. x+7 = x+4 x+x = 4 7 x = 4

25 Appendix 1. Solve the equation x = 1. Solution 1: We have x = 1 x 1 = 0 x 1 = 0 (x+1)(x 1) = 0 x+1 = 0 or x 1 = 0 Solution : We have x = 1 x = 1 x = 1 x = ±1 x = 1 or x = 1 The solutions are x = 1 and x = 1.. Solve the equation x = 4. Solution 1: We have x = 4 x 4 = 0 x = 0 (x+)(x ) = 0 x+ = 0 or x = 0 Solution : We have x = 4 x = 4 x = x = ± x = or x = The solutions are x = and x =.. Solve the equation x = 5. Solution 1: We have x = 5 x 5 = 0 x ( 5) = 0 (x+ 5)(x 5) = 0 x+ 5 = 0 or x 5 = 0 Solution : We have x = 5 x = 5 x = 5 x = ± 5 x = 5 or x = 5 The solutions are x = 5 and x = 5. 5