1. Equations That May be Reduced to Quadratics
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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. 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. 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. 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. 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. 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. 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. 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. 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. c. Given x4 – 5x2 – 14,
Equations That May be Reduced to Quadratics

12. c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
Equations That May be Reduced to Quadratics

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example C. Solve for x.( x
x – 1 )
2
– ( x
x – 1 ) – 6 = 0
Equations That May be Reduced to Quadratics

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. 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. 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. 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. 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. 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. 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. 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. 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. Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
Equations That May be Reduced to Quadratics

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Example E. Solve x + 1
x – 1
– 2 = 3
x + 1
Equations That May be Reduced to Quadratics

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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
)(