1. Expressions

2. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Expressions

3. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”.
Expressions

4. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions

5. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Expressions

6. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
Expressions

7. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
Expressions

8. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
Expressions

9. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100 (pizzas)
8 8
Expressions

10. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100
8 8
“8x + 10 = 810” is called an equation.
(pizzas)

11. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100
8 8
“8x + 10 = 810” is called an equation.
Equations are set up to backtrack to the original
input x or x’s, i.e. we want to solve equations.
(pizzas)

12. Solving Equations

13. An equation is made by setting two expressions
equal to each other:
Solving Equations
expression 1 = expression 2

14. An equation is made by setting two expressions
equal to each other:
Solving Equations
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.

15. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.

16. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.

17. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.

18. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.

19. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
To solve other equations such as rational equations,
or equations with radicals, we have to transform them
into polynomial equations.

20. Rational Equations
Solve rational equations by clearing all denominators
using the LCD.

21. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.

22. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve x – 2
2 = x + 1
4 + 1

23. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
x – 2
2 = x + 1
4 + 1

24. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
p
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
x – 2
2 = x + 1
4 + 1

25. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1)
x – 2
2 = x + 1
4 + 1

26. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2)
x – 2
2 = x + 1
4 + 1

27. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

28. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

29. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

30. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

31. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

32. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
0 = (x + 4)(x – 3)  x = -4, 3
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

33. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
0 = (x + 4)(x – 3)  x = -4, 3
Both are good.
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1

34. Factoring By Grouping
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.

35. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
Factoring By Grouping

36. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
Factoring By Grouping

37. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
Factoring By Grouping

38. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
Factoring By Grouping

39. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping

40. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping

41. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Except for some special cases, polynomial equations
with degree 3 or more are solved with computers.
Factoring By Grouping

42. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
We may also use the quadratic formula to solve all
2nd degree polynomial equations.
Except for some special cases, polynomial equations
with degree 3 or more are solved with computers.
Factoring By Grouping

43. Quadratic Formula and Discriminant
Quadratic Formula (QF):

44. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.

45. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.

46. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots,
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.

47. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.

48. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.

49. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
A “root” is a solution for
the equation “# = 0”.

50. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0,
A “root” is a solution for
the equation “# = 0”.

51. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
A “root” is a solution for
the equation “# = 0”.

52. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
– 4 + 12k > 0
or k > 1/3
A “root” is a solution for
the equation “# = 0”.

53. Power Equations
Equations of the Form xp/q = c

54. Power Equations
The solution to the equation
x 3 = –8
Equations of the Form xp/q = c

55. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8
3
Equations of the Form xp/q = c

56. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Equations of the Form xp/q = c

57. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8
Equations of the Form xp/q = c

58. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2

59. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.

60. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.

61. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.

62. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
then x = (±)c1/R
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
Reciprocate the powers

63. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
then x = (±)c1/R
Reciprocate the powers
The reciprocal of the power 3
xp/q = c
x = (±)cq/por
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.

64. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.

65. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16

66. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Raise both sides to -3/2 power.

67. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2
Raise both sides to -3/2 power.

68. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Raise both sides to -3/2 power.
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

69. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

70. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

71. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

72. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

73. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
2x = 7  x = 7/2
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.

74. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
x-2/3 = 16  (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8  [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
2x = 7  x = 7/2
Since x = 7/2 doesn't work because 43/2 = -8,
there is no solution.

75. Radical Equations

76. Radical Equations
Solve radical equations by squaring both sides to
remove the square root.

77. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2

78. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4

79. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2

80. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4

81. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12

82. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3

83. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2

84. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9

85. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9

86. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
0 = (x – 9)(x – 1)
x = 9, x = 1

87. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
0 = (x – 9)(x – 1)
x = 9, x = 1 Only 9 is good.

88. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|.
Absolute Value Equations

89. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations

90. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{

91. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5.

92. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.

93. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.

94. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
Warning: In general |x ± y|  |x| ± |y|.

95. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
Warning: In general |x ± y|  |x| ± |y|.
For instance, | 2 – 3 |  |2| – |3|  |2| + |3|.

96. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations

97. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.

98. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3

99. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3

100. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then

101. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5

102. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5

103. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|

104. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)

105. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
–4 = x
–4 = x

106. Because |x±y|  |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
–4 = x or 2x – 3 = –3x – 1
–4 = x or x = 2/5

107. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
Absolute Value Equations

108. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations

109. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations
c

110. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations
c
r r

111. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
c
r
x = c + r
r
x = c – r

112. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
c
r
x = c + r
r
x = c – r

113. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
r
x = c – r

114. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
7
1212
r
x = c – r

115. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
So to the right x = 7 + 12 = 19.
and to the left x = 7 – 12 = – 5,
7
1212
x = 19x = – 5
r
x = c – r

116. Given a formula f, the domain of f are all the numbers
that may be computed by f.
Zeroes and Domains

117. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
Zeroes and Domains

118. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
1
x

119. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D

120. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1

121. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1
The zeros of f are where x2 – 1 = 0 or x = ±1.

122. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1
The zeros of f are where x2 – 1 = 0 or x = ±1.
The domain of f are
“all numbers except x2 – 4 = 0, or except x = ±2”.

123. A. Solve the following equations by factoring.
5. x2 – 3x = 10
9. x3 – 2x2 = 0
6. x2 = 4
7. 2x(x – 3) + 4 = 2x – 4
10. 2x2(x – 3) = –4x
8. x(x – 3) + x + 6 = 2x2 + 3x
1. x2 – 3x – 4 = 0 2. x2 – 2x – 15 = 0 3. x2 + 7x + 12 = 0
4. –x2 – 2x + 8 = 0
11. 4x2 = x4
12. 7x2 = –4x3 – 3x 13. 5 = (x + 2)(2x + 1)
14. (x + 1)2 = x2 + (x – 1)2 15. (x + 3)2 – (x + 2)2 = (x + 1)2
B. Solve the following equations by the quadratic formula.
If the answers are not real numbers, just state so.
1. x2 – x + 1 = 0 2. x2 – x – 1 = 0
3. x2 – 3x – 2 = 0 4. x2 – 2x + 3 = 0
5. 2x2 – 3x – 1 = 0 6. 3x2 = 2x + 3
Equations

124. Rational Equations
7. 1
x
+
1
x – 1
=
5
6
8. 1
x
+
1
x + 2
=
3
4
9. 2
x
+ 1
x + 1
= 3
2
10. + 5
x + 2
= 22
x – 1
11. – 1
x + 1
= 3
2
12.6
x + 2
– 4
x + 1
= 11
x – 2
x
6 3
1 2
3
5
2
3–+ = x1. x
4 6
–3 1
8
–5
– 1– = x2.
x
4 5
3 2
10
7
4
3+– = x3. x
8 12
–5 7
16
–5
+ 1+ = x
(x – 20) = x – 3
100
30
100
205.
(x + 5) – 3 = (x – 5)
100
25
100
206.
C. Clear the denominators of the equations, then solve.

125. Radical Equations and Power Equations
D. Isolate one radical if needed, square both sides, do it again
if necessary, to solve for x. Check your answers.
1. x – 2 = x – 4 2. x + 3 = x + 1
3. 2x – 1 = x + 5 4. 4x + 1 – x + 2 = 1
5. x – 2 = x + 3 – 1 6. 3x + 4 = 3 – x – 1
7. 2x + 5 = x + 4 8. 5 – 4x – 3 – x = 1
E. Solve by raising both sides to an appropriate power.
No calculator.
1. x –2 = 1/4 2. x –1/2 = 1/4
3. x –3 = –8 4. x –1/3 = –8
5. x –2/3 = 4 6. x –3/2 = 8
7. x –2/3 = 1/4 8. x –3/2 = – 1/8
9. x 1.5 = 1/27 10. x 1.25 = 32
11. x –1.5 = 27 12. x –1. 25 = 1/32

126. F. Solve for x.
1. Is it always true that I+x| = x? Give reason for your answer.
2. Is it always true that |–x| = x? Give reason for your answer.
Absolute Value Equations
3. |4 – 5x| = 3 4. |3 + 2x| = 7 5. |–2x + 3| = 5
6. |4 – 5x| = –3 7. |2x + 1| – 1= 5 8. 3|2x + 1| – 1= 5
9. |4 – 5x| = |3 + 2x|
11. |4 – 5x| = |2x + 1| 12. |3x + 1| = |5 – x|
10. |–2x + 3|= |3 – 2x|
Solve geometrically for x. Draw the solution.
13. |x – 2| = 1 14. |3 + x| = 5 15. | –9 + x| = –7
x – 4
G. Find the zeros and the domain of the following rational
formulas. (See 2.1)
2x – 1
1. x2 – 1.
x2 – 43.
5x + 7
2.
3x + 5
x2 – x
x2 – x – 24.
x2 – 4x5.
x2 + x – 2
x2 + 2x6.
x2 + x + 2
2x2 – x – 17.
x3 + 2x
x4 – 4x8.
x3 – 8
16. |2 + x| = 1 17. |3 – x| = –5 18. | –9 – x| = 8

127. (Answers to odd problems) Exercise A.
Exercise B.
1. No real solution 3. 𝑥 =
3
2
±
17
2
5. 𝑥 =
1
4
3 ± 17
1. 𝑥 = −1, 𝑥 = 4 3. 𝑥 = −4, 𝑥 = −3 5. 𝑥 = −2, 𝑥 = 5
7. 𝑥 = 2 9. 𝑥 = 0, 𝑥 = 2 11. 𝑥 = ±2, 𝑥 = 0
13. 𝑥 = −3, 𝑥 =
1
2
15. 𝑥 = 2(2 ± 5)
1. 𝑥 =
13
9
3. 𝑥 = 23 5. 𝑥 = 30 7. 𝑥 =
2
5
, 𝑥 = 3
9. 𝑥 =
1
6
3 ± 57 11. 𝑥 =
1
6
5 ± 145
Exercise C.
Equations

128. Exercise D.
1. 𝑥 = 4 3. 𝑥 = 4 5. 𝑥 = 6 7. 𝑥 = 4, 𝑥 =
4
9
Exercise E.
1. 𝑥 = ±2 3. 𝑥 = 2
2
3 5. 𝑥 =
1
8
7. 𝑥 = 8
9. 𝑥 =
1
9
11. 𝑥 =
1
9
1. If 𝑥 < 0, the statement is not true 3. 𝑥 =
1
5
, 𝑥 = 7/5
5. 𝑥 = −1, 𝑥 = 4 7. 𝑥 = 5/2, 𝑥 = −7/2
9. 𝑥 = 1/7, 𝑥 = 7/3 11. 𝑥 =
3
7
, 𝑥 =
5
3
13. |𝑥 − 2| = 1
𝑥 = 1 or 𝑥 = 3
Exercise F.
21 3
left 1 right 1
Radical Equations and Power Equations

129. 15. No solution 17. No solution
Exercise G.
1. Zeros: 𝑥 = 1/2, domain: 𝑥 ≠ 4
3. Zeros: 𝑥 = ±2, domain: 𝑥 ≠ ±4
5. Zeros: 𝑥 = 0, 𝑥 = 4, domain: 𝑥 ≠ −2, 𝑥 ≠ 1
7. Zeros: 𝑥 = −
1
2
, 𝑥 = 1, domain: 𝑥 ≠ 0
Absolute Value Equations