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 ±i2 = 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