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# 5 4 equations that may be reduced to quadratics

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### 5 4 equations that may be reduced to quadratics

1. 1. Equations That May be Reduced to Quadratics Back to 123b homepage
2. 2. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method Equations That May be Reduced to Quadratics
3. 3. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics
4. 4. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler.
5. 5. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given ( x x – 1 ) 2 – ( x x – 1 ) – 6
6. 6. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given if we substitute y for ( x x – 1 ) 2 – ( x x – 1 ) – 6 x x – 1
7. 7. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given if we substitute y for then the expression is y2 – y – 6. ( x x – 1 ) 2 – ( x x – 1 ) – 6 x x – 1
8. 8. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given if we substitute y for then the expression is y2 – y – 6. ( x x – 1 ) 2 – ( x x – 1 ) – 6 x x – 1 b. Given (x2 – 1)2 – 3(x2 – 1) + 2
9. 9. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given if we substitute y for then the expression is y2 – y – 6. ( x x – 1 ) 2 – ( x x – 1 ) – 6 x x – 1 b. Given (x2 – 1)2 – 3(x2 – 1) + 2 if we substitute y for (x2 – 1)
10. 10. In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations Equations That May be Reduced to Quadratics Substitution Method If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler. Example A. a. Given if we substitute y for then the expression is y2 – y – 6. ( x x – 1 ) 2 – ( x x – 1 ) – 6 x x – 1 b. Given (x2 – 1)2 – 3(x2 – 1) + 2 if we substitute y for (x2 – 1) then the expression is y2 – 3y + 2.
11. 11. c. Given x4 – 5x2 – 14, Equations That May be Reduced to Quadratics
12. 12. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, Equations That May be Reduced to Quadratics
13. 13. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics
14. 14. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations.
15. 15. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution,
16. 16. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations.
17. 17. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x.
18. 18. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x. We substitute y = x2 so x4 = y2, the equation is y2 – 5y – 14 = 0 (1st equation to solve)
19. 19. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x. We substitute y = x2 so x4 = y2, the equation is y2 – 5y – 14 = 0 (1st equation to solve) (y – 7)(y + 2) = 0 y = 7, y = –2
20. 20. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x. We substitute y = x2 so x4 = y2, the equation is y2 – 5y – 14 = 0 (1st equation to solve) (y – 7)(y + 2) = 0 y = 7, y = –2 To find x, since y = x2 , so 7 = x2 and –2 = x2
21. 21. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x. We substitute y = x2 so x4 = y2, the equation is y2 – 5y – 14 = 0 (1st equation to solve) (y – 7)(y + 2) = 0 y = 7, y = –2 To find x, since y = x2 , so 7 = x2 and –2 = x2 (2nd equation to solve)
22. 22. c. Given x4 – 5x2 – 14, if we substitute y = x2 so x4 = y2, then the expression is y2 – 5y – 14. Equations That May be Reduced to Quadratics To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations. Example B. Solve x4 – 5x2 – 14 = 0 for x. We substitute y = x2 so x4 = y2, the equation is y2 – 5y – 14 = 0 (1st equation to solve) (y – 7)(y + 2) = 0 y = 7, y = –2 To find x, since y = x2 , so 7 = x2 and –2 = x2 (2nd equation to solve) ±7 = x ±i2 = x
23. 23. Example C. Solve for x.( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 Equations That May be Reduced to Quadratics
24. 24. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 Equations That May be Reduced to Quadratics
25. 25. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 Equations That May be Reduced to Quadratics
26. 26. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
27. 27. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
28. 28. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x 3x – 3 = x ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
29. 29. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x 3x – 3 = x x = 3/2 ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
30. 30. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x –2(x – 1) = x 3x – 3 = x x = 3/2 ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
31. 31. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x –2(x – 1) = x 3x – 3 = x –2x + 2 = x x = 3/2 ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
32. 32. Example C. Solve for x. We substitute y = the equation is y2 – y – 6 = 0 (1st equation to solve) (y – 3)(y + 2) = 0 y = 3, y = –2 To find x, since y = so 3 = and –2 = (2nd equation to solve) 3(x – 1) = x –2(x – 1) = x 3x – 3 = x –2x + 2 = x x = 3/2 2/3 = x ( x x – 1 ) 2 – ( x x – 1 ) – 6 = 0 x x – 1 x x – 1 x x – 1 x x – 1 Equations That May be Reduced to Quadratics
33. 33. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. Equations That May be Reduced to Quadratics
34. 34. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 Equations That May be Reduced to Quadratics
35. 35. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, Equations That May be Reduced to Quadratics
36. 36. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) Equations That May be Reduced to Quadratics
37. 37. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 Equations That May be Reduced to Quadratics
38. 38. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 Equations That May be Reduced to Quadratics
39. 39. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) Equations That May be Reduced to Quadratics
40. 40. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 Equations That May be Reduced to Quadratics
41. 41. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 27/8 = x Equations That May be Reduced to Quadratics
42. 42. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x Equations That May be Reduced to Quadratics
43. 43. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x –8 = x Equations That May be Reduced to Quadratics
44. 44. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x –8 = x Equations That May be Reduced to Quadratics Recall the steps below for solving a rational equation.
45. 45. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x –8 = x Equations That May be Reduced to Quadratics Recall the steps below for solving a rational equation. I. Find the LCD.
46. 46. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x –8 = x Equations That May be Reduced to Quadratics Recall the steps below for solving a rational equation. I. Find the LCD. II. Multiply both sides by the LCD to get an equation without fractions.
47. 47. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x. We substitute y =x1/3 so x2/3 = y2, the equation is 2y2 + y – 6 = 0 (1st equation to solve) (2y – 3)(y + 2) = 0 y = 3/2, y = –2 To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3 (2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3 27/8 = x –8 = x Equations That May be Reduced to Quadratics Recall the steps below for solving a rational equation. I. Find the LCD. II. Multiply both sides by the LCD to get an equation without fractions. III. Solve the equation and check the answers, make sure it doesn't make the denominator 0.
48. 48. Example E. Solve x + 1 x – 1 – 2 = 3 x + 1 Equations That May be Reduced to Quadratics
49. 49. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, x + 1 x – 1 – 2 = 3 x + 1 Equations That May be Reduced to Quadratics
50. 50. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 Equations That May be Reduced to Quadratics
51. 51. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) Equations That May be Reduced to Quadratics
52. 52. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) Equations That May be Reduced to Quadratics
53. 53. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
54. 54. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
55. 55. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
56. 56. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x2 + 2x + 1 – 2x2 + 2 = 3x – 3 x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
57. 57. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x2 + 2x + 1 – 2x2 + 2 = 3x – 3 -x2 +2x + 3 = 3x – 3 x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
58. 58. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x2 + 2x + 1 – 2x2 + 2 = 3x – 3 -x2 +2x + 3 = 3x – 3 0 = x2 + x – 6 x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
59. 59. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x2 + 2x + 1 – 2x2 + 2 = 3x – 3 -x2 +2x + 3 = 3x – 3 0 = x2 + x – 6 0 = (x + 3)(x – 2) x = –3 , x = 2 x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
60. 60. Example E. Solve The LCD is (x – 1)(x + 1), multiply the LCD to both sides, (x – 1)(x + 1) [ ] (x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1) x2 + 2x +1 – 2(x2 – 1) = 3x – 3 x2 + 2x + 1 – 2x2 + 2 = 3x – 3 -x2 +2x + 3 = 3x – 3 0 = x2 + x – 6 0 = (x + 3)(x – 2) x = –3 , x = 2 Both solutions are good. x + 1 x – 1 – 2 = 3 x + 1 x + 1 x – 1 – 2 = 3 x + 1 (x + 1) (x – 1)(x + 1) ( x – 1) Equations That May be Reduced to Quadratics
61. 61. Equations That May be Reduced to Quadratics Exercise A. Solve the following equations. Find the exact and the approximate values. If the solution is not real, state so. 1. x4 – 5x2 + 4 = 0 3. 2x4 + x2 – 6 = 0 2. x4 – 13x2 + 36 = 0 4. 3x4 + x2 – 2 = 0 5. 2x4 + 3x2 – 2 = 0 6. 3x4 – 5x2 – 2 = 0 7. 2x6 + x3 – 6 = 0 8. 3x6 + x3 – 2 = 0 9. 2x6 + 3x3 – 2 = 0 10. 3x6 – 5x3 – 2 = 0 11. 2x + x1/2 – 6 = 0 12. 3x + x1/2 – 2 = 0 13. 2x + 3x1/2 – 2 = 0 14. 3x – 5x1/2 – 2 = 0 15. 2x –2/3 + x –1/3 – 6 = 0 16. 3x –2/3 + x – 1/3 – 2 = 0 17. 2x –2/3 + 3x – 1/3 – 2 = 0 18. 3x –2/3 – 5x – 1/3 – 2 = 0 19. 2 2 + – 6 = 0 2x – 3 x + 1)(2x – 3 x + 1)( 20. 2 –2 + –1 – 6 = 0 x + 1 )(2x – 3 2x – 3 x + 1 )(