5 4 equations that may be reduced to quadratics

792 views

Published on

Published in: Technology, Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
792
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
0
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

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 )(

×