Intermediate Algebra Review
Intermediate Algebra Review
Evaluation of algebraic expressions by substitution
Algebraic expressions contain letters which are referred to as variables. If we replace a variable
with a number, it is referred to as substitution. Then, when we carry out any calculations, it is
called evaluating the expression.
Example 1: Find the value of ac when a=15 and c=5.
Substitute 15 for a and 5 for c and carry out the multiplication.
ac=(15)(5) =75
Example 2: m=20 and n=4. Find the value of ½(m-n).
In the expression ½(m-n), replace m with 20 and n with 4. Then carry out the subtraction
and multiplication.
1
1
1
(m-n) = 2 (20-4) = 2(16) = 8
2
Example 3: Evaluate the expression P =
𝑃−2𝐿
2
when P = 90 and L = 26.
Replace P with 90 and L with 26. Simplify the numerator and divide by 2.
𝑃−2𝐿
P=
2
=
90−2𝑥26
2
=
90−52
2
=
38
2
= 19
Here are some problems for you to try. The answers are at the bottom of the page.
Evaluate the following:
1. 2m2 when m = 8
𝑢
2. uv+𝑣 when u=12 and v=4
3. k(k+h) when k=6 and h=7
4. 3p-2s when p=9 and s=5
5. a-bc when a=12, b=6, and c=2
6.
𝑟+𝑡
2
when r=15 and t=9
7. (𝑒 + 𝑓)2 when e=8 and f=4
8. 𝑒 + 𝑓 2 when e=8 and f=4
9. 𝑒2+f when e=8 and f=4
10. (𝑤 + 𝑦)(𝑤 − 𝑦)when w=5 and y=4
𝑝+𝑞
11. 𝑝−𝑞 when p=10 and q=5
Multiplication symbols: *
● or x
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12. (𝑏 − 𝑐)2when b=7 and c=4
13. x2 when x= -3
14. –x2 when x = -3
Answers
1. 128
2. 51
3. 78
4. 17
5. 0
6. 12
7. 144
8. 24
9. 68
10. 9
11. 3
12. 9
13. 9
14. -9
Addition, subtraction, and multiplication of polynomials
When two or more terms have the same variables raised to the same powers, they are considered
like terms. Like terms can be added or subtracted. Likewise, polynomials can be added or
subtracted by “collecting” or “combining” like terms.
Example 1: Collect like terms
3a2 – 4b + 2a2 =5a2 – 4b
Example 2: Add
(-6v3 + 2v – 4) + (4v3 + 3v2 +2) = -2v3 + 3v2 +2v -2
Example 3: Subtract
(4p4q – 5p3q2 + p2q3 +2q4) – (-5p4q + 5p3q2 – 3p2q3 – 7q4) = 9p4q – 10p3q2+ 4p2q3 + 9q4
Here are some problems for you to try. The answers follow the exercise.
Collect like terms
1. 3y – 4x + 6xy2 – 2xy2
Multiplication symbols: *
● or x
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2. 3ab3 + 2a3b + 5ab3 -8a +15 – 3a2b – 6a2b + 11a -8
Add.
3.
4.
5.
6.
(7x5 -5) + (3x5 -4x2 +10)
(5a2b4 – 2a2b2 – 3b) + -6a2b2 +3b +5)
(4ax2 + 4 bx -5) + (-6ax2 + 8)
(13g3h +3g2h -5h) + (g3h + 4g2h – 3gh)
Subtract
7. (-5a2 +4) – (2a2 + 3a – 1)
8. (9x3-2x – 4) – (-5x3-8)
9. (2x5 – x4 +3x3 –x2 – x – 7) – (-x5 +2x4 -2x3 +x2 – x – 4)
10. (4x4y – 5x3y2 + x2y3 +2y4) – (-5x4y + 5x3y2 – 3x2y3 – 7y4)
Answers
1. 3y – 4x + 4xy2
2. 8ab3 + 2a3b – 9a2b + 3a + 7
3. 10x5 – 4x2 + 5
4. 5a2b4 – 8a2b2 + 5
5. -2ax2 + 4bx + 3
6. 14g3h + 7g2h – 3gh –5h
7. -7a2 – 3a +5
8. 14x3 – 2x + 4
9. 3x5 – 3x4 + 5x3 –2x2 – 3
10. 9x4y – 10x3y2 + 4x2y3 + 9y4
Solution of linear equations in 1 variable
Linear equations in 1 variable are solved using the addition and multiplication principles. The
solution determines the replacement for x which will make the equation true.
Example 1:
Example 2:
x + 6 = -15
check:
x + 6 = -15
x + 6 + (-6) = -15 + (-6)
-21 + 6 = -15
x = -21
-15 = -15
4
4
𝑥 = 22
5
5
4
4
5
∗ 5x = 4 ∗
x=
Multiplication symbols: *
55
2
check: 5 𝑥 = 22
22
4
1
5
∗
55
2
= 22
22=22
● or x
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Here are some problems for you to try.
Solve.
1.
2.
3.
4.
5.
6.
7.
8x = 10
−3
𝑥 = 21
7
−𝑥 = −5
−𝑥
= 17
8
−4𝑥 + 2 + 5𝑥 = 3𝑥 − 15
30 + 7(𝑥 −1) =3(2x + 7)
3
2
𝑥 + 5 = 3𝑥 − 7
4
Answers
1.
2.
3.
4.
5.
6.
7.
5
4
-49
5
-136
17
2
-2
-144
Solution of linear inequalities in 1 variable
Inequalities are solved using the same principles that are used for solving equalities. There is
one difference: When solving an inequality, if you multiply or divide by a negative number, the
inequality sign changes direction.
Example 1:
3–x<2
-x< -1
x>1
Example 2:
2m – 22 > 16 (m – 4)
2m – 22 > 16m – 64
-14m > -42
m<3
Here are some practice problems for you to try
1.
2.
3.
4.
x – 2 >6
3x + 5 < -10
x – 6 > 2x – 5
5t – 3 < 9 – t
Multiplication symbols: *
● or x
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5. 0.6z < -18
6. 2c – 7 < 5c – 9
7. 4(4x – 3) > 9(2x + 7)
8. 3(2 – 5x) + 2x < 2(4 + 2x)
9. 5(t + 3) + 9 < 3(t-2) +6
10. 13 – (2c +2) > 2(c + 2) + 3c
Answers
1. x > 8
2. x < -5
3. x < -1
4. t < 2
5. z < -30
6. c > 2/3
7. x < -75/2
8. x > -2/17
9. t < -12
10. c < 1
Factorization of polynomials
Factoring is the process of “unmultiplying,” that is, it is the opposite of multiplying. When you
factor a term or an expression, you find an equivalent term or expression which is a product.
Sometimes you will only be able to factor out a common term or the “greatest common factor.”
Other times you will be looking for two or more binomials. Sometimes you will need to do both.
When you multiply a monomial and polynomial, you multiply the monomial by the polynomial
using the distributive property of multiplication. When you factor, you do the opposite.
Example 1:
Example 2:
Multiply
Factor
3x(6x2 + 2x – 1)
18x3 + 6x2 -3x
= 18x3 + 6x2 -3x
= 3x(6x2 + 2x -1)
Factor
10x6y2 – 4x5y3 + 2x4y4 – 2x4y2 = 2x4y2(5x2 - 2xy + y2 -1)
When you multiply two binomials, you use the FOIL method.
FOIL
(x + 6)(x – 4) = x2 – 4x + 6x – 24 = x2 + 2x – 24
and when you factor a trinomial you are looking for the two binomials whose product is the
trinomial.
Example 3:
x2 + 2x – 24 = (x + 6)(x – 4)
Multiplication symbols: *
● or x
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Example 4:
3x2 + 19x + 20 = (3x + 4)(x + 5)
Sometimes you must first take out the greatest common factor and then find the two binomials:
Example 5:
20x5 – 46x4 + 24x3 = 2x3 (10x2 -23x + 12) = 2x3 (2x – 3)(5x – 4)
Here are some practice problems.
Factor
1.
2.
3.
4.
5.
6.
x2 – 14x +45
3y2 – 20y +32
9y2 + 15y + 4
14x4 – 19x3 –3x2
5x + x2 –14
56x + x2 – x3
Answers
1.
2.
3.
4.
5.
6.
(x – 9)(x – 5)
(3y – 8)(y – 4)
(3y + 1)( 3y + 4)
x2(7x+1)(2x – 3)
(x + 7)(x – 2)
x(8 – x)(7 + x)
There are also several “special” types of factorization with which you should be familiar.
A trinomial square factors into two binomials which are identical:
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Example 1:
x2 – 4x + 4 = (x – 2)(x – 2) = (x – 2)2
Example 2:
9y2 + 12y + 4 = (3y + 2)(3y + 2) = (3y + 2)2
These problems are trinomial squares. Try them using the pattern above.
Factor.
1.
2.
3.
4.
x2 + 14x + 49
9x2 – 30x + 25
16x2 + 72xy +81y2
-8a2 + 24ab – 18b2
Answers
1. (x + 7)(x + 7) = (x + 7)2
2. (3x – 5)(3x – 5) = (3x – 5)2
Multiplication symbols: *
● or x
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3. (4x + 9y)(4x + 9y) = (4x + 9y)2
4. -2(2a – 3b)(2a – 3b) = -2(2a – 3b)2
These are examples of differences of squares:
y2 – 9
x4 – 4y4
9a2 – 36b2
And this is the pattern for factoring the difference of two squares:
A2 – B2 = (A + B)(A – B)
Example 1:
y2 – 16 = (y + 4)(y – 4)
Example 2:
4x3 – 49x = x(2x + 7)(2x – 7)
Try these problems using the pattern.
Factor.
1.
2.
3.
4.
5.
x2 – 9
9y2 – 36
2x4 – 8y4
20x2 – 5y2
a4 – 16b4
Answers
1.
2.
3.
4.
5.
(x + 3)(x – 3)
9(y + 2)(y – 2)
2(x2 – 2y2)(x2 + 2y2)
5(2x – y)(2x + y)
(a – 2b)(a + 2b)(a2 +4b2)
Another “special” form of factorization involves factoring sums or differences of cubes. Here
are the patterns that must be used:
A3 + B3 = (A + B)(A2 – AB + B2)
A3 – B3 = (A – B)(A2 + AB + B2)
Example 1:
x3 – 8 = (x3 – 23) = (x – 2)(x2 + 2x + 22) = (x – 2)(x2 + 2x + 4)
Example 2:
24a3 + 3 = 3(8a3 +1) = 3[(2a)3 + 13]
= 3(2a + 1)[(2a)2 – (2a)(1) + 12] = 3(2a + 1)(4a2 – 2a +1)
Here’s some practice for you.
Factor
1. a3 + 27
2. 8 – 27b3
Multiplication symbols: *
● or x
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3. 64x3 + 1
4. ab3 + 125a
Answers
1.
2.
3.
4.
(a + 3)(a2 – 3a +9)
(2 – 3b)(4 + 6b +9b2)
(4x +1)(16x2 – 4x +1)
a(b + 5)(b2 – 5b + 25)
Solution of polynomial equations by factoring
Now that you have reviewed factoring polynomials, you should be ready to solve quadratic
equations. Follow these steps to solve such an equation:
1)
2)
3)
4)
Move all the terms to one side of the equal sign so the equation is equal to zero.
Factor completely
Set each facto equal to zero
Solve the equations
Example 1:
7x + 3x2 = -2
3x2 +7x +2 =0
(3x +1)(x + 2) = 0
3x +1 =0
1
x=-3
Example 2:
x+2=0
x = -2
3x3 – 9x2 = 30x
3x3 – 9x2 – 30x = 0
3x(x2 – 3x – 10) = 0
3x(x - 5)(x + 2) = 0
3x = 0
x–5=0
x+2=0
x=0
x=5
x = -2
Here are some practice problems for you.
1.
2.
3.
4.
5.
6.
7.
x2 + 8 = 6x
5y + 2y2 = 3
8b2 = 16b
25 + x2 = -10x
x3 + x2 = 6x
9y2 + 15y + 4 = 0
27 + 12t + t2 = 0
Multiplication symbols: *
● or x
November 2012 Document1
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8. 8y – y2 = 0
9. 2x3 – 2x2 =12x
10. 2x3 = 128x
Answers
1. 2, 4
2. -3, ½
3. 0, 2
4. -5
5. -3, 0, 2
6. -4/3, -1/3
7. -9, -3
8. 0, 8
9. -2, 0, 3
10. -8, 0, 8
Properties of square roots
The processes of “squaring” and “finding square roots” are opposite, or inverse, operations.
Example 1:
Square
Square root
32 = 9
√9 = 3
72 = 49
√49 = 7
You can also find square roots of expressions which contain variables. (Note: Throughout this
packet assume all variables under radical signs represent positive numbers.)
Example 2:
Square
Square root
(3x)2 = 9x2
√9𝑥2 = √(3𝑥)2 = 3x
(ab)2 = a2b2
√𝑎2 𝑏 2 = √(𝑎𝑏)2 =ab
Try these problems.
Simplify
1.
2.
3.
4.
5.
6.
7.
8.
√225
√441
√(7𝑤)2
√(𝑥𝑦)2
√𝑥 2 𝑦 2
√(𝑥 − 11)2
√25𝑦 2
√𝑥 2 + 8𝑥 + 16
Multiplication symbols: *
● or x
November 2012 Document1
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9. -√81
1
10. √4 𝑎2
Answers
1. 15
2. 21
3. 7w
4. xy
5. xy
6. x-11
7. 5y
8. x + 4
9. -9
10. 1/2a
Rational expressions
1 −5 16
50
2
1
,
8
, −3 , 𝑎𝑛𝑑
x2 + 6x + 5
x2 – 3x +2
are examples of rational numbers
and y2 – 9 are rational expressions or fractional expressions.
y+3
Because rational expression indicate division, you need to remember that a denominator (the
bottom of the fraction) of zero makes the expression undefined.
Look at the examples below:
x2 + 6x + 5 = (x+5)(x+1)
x2 – 3x + 2 (x-2)(x-1)
In this expression, replacing x with either 2 or 1 would create a
denominator a zero. Therefore, we say it is undefined for x=1 or 2.
y2 – 9
y+3
This expression is undefined for y = -3, because replacing y with
-3 would create a zero in the denominator.
Recall that a numerator (the top of the fraction) of zero does not make the expression undefined.
That is because if you divide a number into zero, the answer is zero – and zero is perfectly
acceptable as an answer!
Here are some practice problems. Remember, you may need to factor the denominator before
you can determine your answer.
Find all numbers for which the rational expressions are undefined.
1.
6
2𝑥
Multiplication symbols: *
● or x
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2.
𝑥−6
𝑥+8
3. x2 – 4y + 9
2y + 5
4. x2 – 9
x2 – 7x + 10
Answers
1. 0
2. -8
−5
3. 2
4. 2, 5
Multiplying rational expressions is just like multiplying fractions. Multiply the numerators to get
the numerator of the product and multiply the denominators to get the denominator of the
product. When multiplying rational expression, always factor both the top and bottom to reduce
the fractions, if possible, before you multiply.
(3𝑥+2𝑦)𝑥
3𝑥+2𝑦
𝑥
𝑥+2
𝑥 2 −4
3𝑥 2 +2𝑥𝑦
● = (5𝑥+4𝑦)𝑥 = 5𝑥 2 +4𝑥𝑦
5𝑥+4𝑦 𝑥
Example 1:
1
Example 2:
𝑥+2
(𝑥+2)(𝑥−2)
(𝑥+2)(𝑥−2)
● 𝑥 2 +𝑥−2 = 𝑥−3 ● (𝑥+2)(𝑥−1) = (𝑥−3)(𝑥−1)
𝑥−3
1
Dividing rational expressions is just like multiplying except you must first invert (flip over) the
second expression. The problem then becomes a multiplication.
𝑥−2
Example 3:
Example 4:
𝑥+1
12𝑥 8
3𝑦 4
𝑥+5
𝑥−2
𝑥−3
3𝑥 5
2
(𝑥−2)(𝑥−3)
÷ 𝑥−3 = 𝑥+1 ● 𝑥+5 = (𝑥+1)(𝑥+5)
÷
16𝑥 3
=
6𝑦
12𝑥 8
3𝑦 4
6𝑦
3𝑥 5
● 16𝑥 3 = 2𝑦 3
y3
2
1
𝑦 2 −36
3𝑦−18
1
(𝑦+6)(𝑦−6)
Example 5: 𝑦 2 −8𝑦+16 ÷ 𝑦 2 −𝑦−12 = (𝑦−4)(𝑦−4) ●
1
(𝑦−4)(𝑦+3)
3(𝑦−6)
=
(𝑦+6)(𝑦+3)
3(𝑦−4)
1
Try these practice problems before you go on to adding and subtracting rational expressions.
Multiply or divide as indicated. Simplify if possible.
1.
2.
2𝑥 2 + 20x + 50
𝑥 2 −4
𝑥 2 −16
2𝑥+6
𝑥+2
● 𝑥+5
𝑥−4
÷ 𝑥+3
Multiplication symbols: *
● or x
November 2012 Document1
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3.
4.
5.
6.
𝑥 3 −8
●
𝑥 2 −25
9𝑐 2 −1
𝑐 2 −9
𝑦 2 −64
2𝑦+10
𝑥 3 −64
𝑥 2 −16
÷
𝑥 2 +10𝑥+25
𝑥 2 +2𝑥+4
3𝑐+1
𝑐+3
𝑦+5
● 𝑦+8
𝑥 2 +5𝑥+6
÷ 𝑥 2 −3𝑥−18
Answers
1. 2(x + 5)
x−2
2. x + 4
2
3. (x− 2)(x + 5)
x-5
4. 3c – 1
c–3
5. y – 8
2
6. (x – 6)(x2 + 4x + 16)
(x + 4)(x + 2)
Adding and subtracting rational expressions follows the same rules as adding and subtracting
fractions.
If the expressions have the same denominator, add or subtract the numerators and keep the
same denominator.
3+𝑥
Example 1:
𝑥
4𝑥+5
Example 2:
𝑥+3
4
+𝑥=
3+𝑥+4
𝑥
𝑥−2
− 𝑥+3 =
=
7+𝑥
𝑥
4𝑥+5−(𝑥−2)
𝑥+3
=
4𝑥+5−𝑥+2
𝑥+3
=
3𝑥+7
𝑥+3
As with fractions, you should reduce your answer to lowest terms whenever possible.
1
4𝑥 2 −5𝑥𝑦
Example 3:
𝑥 2 −𝑦2
+
2𝑥𝑦− 𝑦2
𝑥 2 −𝑦 2
=
4𝑥 2 −3𝑥𝑦−𝑦 2
𝑥 2 −𝑦 2
=
(4𝑥+𝑦)(𝑥−𝑦)
(𝑥+𝑦)(𝑥−𝑦)
=
4𝑥+𝑦
𝑥+𝑦
1
If the expressions have different denominators, first find the least common denominator
(LCD) and then add or subtract the numerators. This sometimes requires factoring.
Example 4:
2𝑎
5
Multiplication symbols: *
3𝑏
+ 2𝑎 =
2𝑎
5
×
2𝑎
+
2𝑎
3𝑏
5
× 5 [LCD = (5)(2a) = 10a]
2𝑎
● or x
November 2012 Document1
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4𝑎2 15𝑏 4𝑎2 + 15𝑏
+
=
10𝑎 10𝑎
10𝑎
2𝑦+1
Example 5:𝑦 2 −7𝑦+6 −
(2𝑦+1)
(𝑦−6)(𝑦−1)
2𝑦 2 +3𝑦+1−𝑦 2 −2𝑦+3
=
(𝑦−6)(𝑦−1)(𝑦−1)
𝑦+3
𝑦 2 −5𝑦−6
𝑥
(𝑦+1)
(𝑦+1)
=
2𝑦+1
(𝑦−6)(𝑦−1)
(𝑦+3)
− (𝑦−6)(𝑦+1 𝑥
−
𝑦+3
(𝑦−6)(𝑦+1)
(𝑦−1)
(𝑦−1)
=
[LCD = (y – 6)(y – 1)(y + 1)]
(2𝑦+1)(𝑦+1)−(𝑦+3)(𝑦−1)
(𝑦−6)(𝑦+1)(𝑦−1)
=
𝑦 2 +𝑦+4
(𝑦−6)(𝑦+1)(𝑦−1)
Try these problems on your own.
Add or subtract as indicated. Check each answer to see if it can be reduced.
1.
2.
3.
4.
5.
6.
7.
8.
9.
𝑥−2𝑦
𝑥+𝑦
+
4𝑎−2
𝑦−2
2𝑦−3
𝑥+𝑦
5+3𝑎
𝑎2 −49
4𝑦+3
𝑥+9𝑦
+ 49−𝑎2
−
𝑦−2
𝑦−2
4−𝑦
𝑦 2 −1
− 1−𝑦 2
𝑦−2
𝑦+3
𝑦+4
𝑥−2
𝑥+3
+ 𝑦−5
𝑥+2
+
𝑥−4
9𝑥+2
3𝑥 2 −2𝑥−8
𝑥−1
3𝑥+15
−
3𝑦+2
𝑦 2 +5𝑦−24
+
7
3𝑥 2 +𝑥−4
𝑥+3
5𝑥+25
+
3𝑥−1
7
𝑦 2 +4𝑦−32
𝑥+4
10. 𝑥 2 +2𝑥−3 − 𝑥 2 −9
Multiplication symbols: *
● or x
November 2012 Document1
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Answers:
2𝑥+7𝑦
1. 𝑥+𝑦
2.
3.
4.
5.
6.
7.
8.
9.
1
𝑎+7
3𝑦+5
𝑦−2
1
𝑦−1
2𝑦 2 +22
(𝑦+4)(𝑦−5)
−11𝑥+2
(𝑥+3)(𝑥−4)
3𝑥−4
(𝑥−2)(𝑥−1)
2𝑥−14
15(𝑥+5)
3𝑦 2 −3𝑦−29
(𝑦+8)(𝑦−3)(𝑦−4)
2𝑥 2 −13𝑥+7
10. (𝑥+3)(𝑥−1)(𝑥−3)
Radical Expressions
Earlier in this packet you reviewed square roots. Hopefully, you understand that there are many
roots other than square roots – cube roots, 4th roots, 5th roots – an infinite number of possibilities.
First we will review how to simplify radicals.
To simplify a radical expression, you first must look for factors of the radicand (the number
3
3
under the radical sign) that are perfect powers. [For instance, √16 = √8 ∗ 2 where the number 8
is a perfect cube (23 = 8).} Then take the appropriate root of the resulting factors. The radical
expression is simplified when its radicand has no factors that are perfect powers. This is more
easily seen through examples.
Example 1:
Example 2:
√50 = √25 ∗ 2 = √25 *√2 = 5√2
4
4
4
4
4
√243 = √81 ∗ 3 = √81 *√3 = 3√3
Multiplication symbols: *
● or x
November 2012 Document1
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Example 3:
3
√6𝑥 5 = 3√6𝑥 3 𝑥 2 = 3√𝑥 3 * 3√6𝑥 2 = x * 3√6𝑥 2
Try these practice problems.
Simplify the following radical expressions by factoring.
1. √24
2. √𝑥 4
3
3. √54
√54𝑥 6
4.
3
5.
4
6.
4
7.
3
√(𝑥 + 𝑦)4
√(𝑥 + 𝑦)6
√−24𝑥 4 𝑦 5
Answers:
1. 2√6
2. x2
3
3. 3 √2
3
4. 3x2 ● √2
5. x + y
6. (x + y) ● 4√(𝑥 + 𝑦)2
7. -2xy ● 3√3𝑥𝑦 2
Multiplying radical expression is not difficult: simply multiply the radicands and simplify, if
possible.
Example 1:
√3 * √6 = √18 = √9 ∗ 2 = 3√2
Example 2:
3√25 ● 2√5 = 6 √125 = 6 ● 5 = 30
Example 3:
3
3
3
3
√2 ∗ (𝑥 + 5) * 3√4 ∗ (𝑥 + 5)4 = 3√8 ∗ (𝑥 + 5)5 = 2 * (x + 5) * 3√(𝑥 + 5)2
Dividing follows the same principles. Divide the radicands and simplify if you can.
Example 4:
√80
80
√5
5
Multiplication symbols: *
=√
= √16 = 4
● or x
November 2012 Document1
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3∗ √2
Example 5:
5∗ √3
=
3∗ √2
5∗ √3
∗
(√3)
(√3)
=
3 √6
15
=
√6
5
(This is called rationalizing the
denominator)
Here are some practice problems.
Multiply or divide as indicated. Simplify if possible.
1. √2 ● √32
3
3
2. √3 ● √18
3
3.
4.
5.
6.
7.
8.
5 √32
3
√2
√72𝑥𝑦
2∗ √2
√52 ∗ 𝑡 4 * 3√54 ∗ 𝑡 6
3
√𝑥 3 − 𝑦3
(hint: factor x3 – y3)
√𝑥−𝑦
5
√5
3√6
2√12
Answers
1. 8
3
2. 3 ● √2
3
3. 10 ● √2
4. 3√𝑥𝑦
3
5. 25t3 ● √𝑡
6. √𝑥 2 + 𝑥𝑦 + 𝑦 2
7. √5
3√2
8. 4
When you add or subtract radical expression, sometimes it is not possible to simplify the answer.
For instance, 3 + √5 cannot be simplified. However, if two radicals have the same radicand and
the same index (the small raised number that indicates the root), you can simplify by collecting
the terms.
Multiplication symbols: *
● or x
November 2012 Document1
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Example 1:
7 ● √3 + √3 = 7 ● √3 + 1●√3 = 8 ●√3
Example 2:
3 ● √8 - 5√2 = 3 ● √4 ∗ 2 – 5 √2 = 3 ● 2 * √2 – 5 ● √2 = 6 ● √2 – 5 ● √2 = √2
Example 3: 5 ● 3√16𝑦 4 + 7 ● 3√2𝑦 = 5 ● 3√8 ∗ 2𝑦 3 + 𝑦 + 7 3√2𝑦 = 5 ● 2 ● y● 3√2𝑦 +
7 3√2𝑦 = 10y ● 3√2𝑦 + 7 ● 3√2𝑦 = (10y + 7) ● 3√2𝑦
Practice problems.
Add or subtract. Simplify if possible.
1. 5 ●√2 + 8 ● √2
4
4
2. 7 ● √5𝑥 + 3 ● √5𝑥 - 7
3. 7 ● √45 – 2 ●√5
4. 3 ● 3√𝑦 5 + 4 ● 3√𝑦 2 + 3√8𝑦 6
5. √25𝑥 − 25 − √9𝑥 − 9
6. 2 ●√128 − √18 + 4 ● √32
3
3
3
7. 5 ● √32 − √108 + 2 ● √256
8. √𝑥 3 − 𝑥 2 + √9𝑥 − 9
Answers
1. 13 ● √2
4
2. 10 ● √5𝑥 – 7
3. 19 ● √5
4. (3y + 4) ● 3√𝑦 2 + 2y2
5. 2 ● √𝑥 − 1
6. 29 ● √2
3
7. 15 ● √4
8. (x + 3) ● √𝑥 − 1
Quadratic formula
Earlier you reviewed solving quadratic equations by factoring and applying the principal of zero
products. Quadratic equations can also be solved using the quadratic formula:
Multiplication symbols: *
● or x
November 2012 Document1
18
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
To use the formula you must first put the equation into standard form:
ax2 + bx + c = 0
For example, the equation 5x2 + 8x = -3 becomes 5x2 + 8x + 3 = 0 and a = 5, b = 8, and c = 3.
Now substitute these values into the quadratic formula and solve for x.
𝑥=
x=
−8±√82 −4(5)(3)
2(5)
−8+2
10
=
−3
5
=
−8+ √4
or x =
10
−8−2
10
= -1
Here’s another example. Solve 3x2 = 18x − 6 for x
3x2 = 18x – 6
3x2 – 18x + 6 = 0 so a = 3, b = -18 , c = 6
𝑥=
−(−18)±√(−18)2 −4(3)(6)
2(3)
=
18±√324−72
6
=
18±√252
6
=
18±√36∗7
6
=
18±6√7
6
= 3 ± √7
So
x = 3 + √7 or x = 3 − √7
Practice problems
Solve for x using the quadratic equation.
1. x2 + 6x + 4 = 0
2. x2 – 6x – 4 = 0
3. 3x2 = -8x −1
4. h2 + 4 = 6h
5. 15x2 = 17x – 2
6. 4x + x(x−3) = 0
Answers
1. -3 ± √5
2. 3 ± √13
Multiplication symbols: *
● or x
November 2012 Document1
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3.
−4
3
±
√13
3
4. 3 ± √5
2
5. 15 , 1
6. −1, 0
Numbers to fractional powers
Fractional exponents are another way of writing radical expressions. In a fractional exponent,
the denominator of the fraction indicates the root or index. The numerator is the exponent of the
base.
Example 1:
x1/2 = √𝑥
Example 2:
(xy)1/4 = 4√𝑥𝑦
Example 3:
5
Example 4:
8 ● 3√(𝑥𝑦)2 = 8 (xy)2/3
√7𝑥𝑦 = (7xy)1/5
These kinds of expressions can often be simplified.
3
(27)2/3 = 3√272 = (√27)2 = 32 = 9
Example 5:
Rewrite the following without rational exponents. That is, rewrite using radicals.
Simplify if possible.
1. y1/4
2. x2/3
3. 45/2
4. (125)1/3
5. (a3b2c)1/5
6. 43/2
Rewrite the following with rational exponents
3
7. √19𝑎𝑏
3
8. 19 ● √𝑎𝑏
Multiplication symbols: *
● or x
November 2012 Document1
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5
𝑥2𝑦
9. √ 16
10. 5√𝑎5 𝑏10
Answers
1. 4√𝑦
3
2. √𝑥 2
3. 32
4. 5
5.
√𝑎3 𝑏 2 𝑐
5
6. 8
7. (19ab)1/3
8. 19(ab)1/3
𝑥2𝑦
9. ( 16 )1/5 =
𝑥 2/5 𝑦 1/5
161/5
10. a5/5b10/5 = ab2
Systems of linear equations in 2 variables
The solution of a system of two linear equations in two variables is the ordered pair of numbers
which makes both statements true.
Example 1:
x+y=3
The order pair (-4, 7), when substituted into these two equations,
makes both statements true.
5x – y = -27
x+y=3
-4 + 7 = 3 (true)
5x – y = -27
5(-4) – 7 = -27
-20 – 7 = -27 (true)
The ordered pair (-4, 7) can also be described by saying that x = -4 and y = 7.
Systems of equations may be solved a number of ways. One way is to graph both equations.
The point of intersection, if one exists, is the solution to the system.
Multiplication symbols: *
● or x
November 2012 Document1
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Example 2:
1) y – x = 1 is the same as y = x + 1
2) y + x = 3 is the same as y = -x + 3
1
1) slope = 1
y intercept = 1
−1
2) slope = 1
y intercept = 3
Solution is (1, 2)
Note: graphing is not always accurate.
Note: If the lines are parallel, there will be no point of intersection and no solution to the system
of equations. If both equations graph as the same line, there are an infinite number of solutions
to the system of equations.
There are also other ways to solve a system of equations. One of those is the substitution
method.
Example 3:
2) x + y = 4
3) x = y +1
In this example you can substitute y + 1 for the x in the first equation.
1) x + y = 4
1) y + 1 + y = 4
2y + 1 = 4
2y = 3
3
y=2
3
Next, replace the y in either equation with 2; then solve for x.
x+y=4
x=y+1
3
3
x+2 =4
x =2 + 1
x=
5
x=
2
5
2
You may also use the elimination method which involves the addition principle. In this method,
you create—by multiplication—coefficients which are equal but which have opposite signs.
You can then add the two equations together to eliminate one of the variables and solve for the
other.
Example 4:
2x – 3y = 0
-4x + 3y = -1
Multiplication symbols: *
In this problem the coefficients of y are already equal but opposite.
Add the two equations to eliminate the y.
● or x
November 2012 Document1
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-2x + 0y = -1
-2x = -1
1
x=2
1
Now replace the x in either equation with 2; then solve for y.
2x – 3y = 0
1
2(2) – 3y = 0
1 – 3y = 0
-3y = -1
1
y=3
-4x + 3y = -1
1
-4(2) + 3y = -1
-2 + 3y = -1
3y = 1
1
y= 3
Example 5:
3x + 3y = 15
2x + 6y = 22
Multiply the first equation by -2; then add it to the second.
-6x + -6y = -30
2x + 6y = 22
-4x + 0y = -8
x=2
Now replace x with 2 in either equation and solve for y.
3x + 3y = 15
3(2) + 3y =15
6 + 3y = 15
3y = 9
y=3
2x + 6y = 22
2(2) + 6y = 22
4 + 6y = 22
6y = 18
y=3
Now you can try some yourself.
Solve the following systems of equation using any of the three methods discussed above.
1. x + y = 6
y=x+2
2. 8x + 5y = 184
x – y = -3
Multiplication symbols: *
● or x
November 2012 Document1
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3. 5x + 3y = 17
-5x + 2y = 3
4. 4x + y = -1
x – 2y = 11
5. 5m + n = 8
3m – 4n = 14
6. 5r – 3s = 24
3r + 5s = 28
Answers:
1. x = 2, y = 4 2. x = 13, y = 16
1. m = 2, n = -2 6. r = 6, s = 2
3. x = 1, y = 4
4. x = 1, y = -5
Absolute value equations
To solve equations involving absolute value, you must recall the absolute value principle:
a) For any positive number p, if |𝑥| = p, then x = -p or x = p.
b) For the equation |𝑥| = 0, the solution is x = 0.
c) |𝑥| = n
has no solution when n is a negative number
Here are examples of the preceding principle.
Example 1:
|𝑥| = 4
x = 4 or x = -4
This follows from the fact that |4| = 4 and |−4| = 4.
Example 2:
Example 3:
|2 + 5𝑥| = 13
2x + 5 = 13
2x = 8
x=4
or
2x + 5 = -13
2x = -18
x = -9
|𝑥| = -15 has no solution
This is because absolute value is by definition positive.
Practice on the following problems
1. |𝑥| = 6
2. |𝑥| = 0
3. |𝑥| = -8
Multiplication symbols: *
● or x
November 2012 Document1
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4. |𝑥 − 4| = 1
5. |3𝑥| = 6
6. |3𝑥 − 4| = 17
7. |2𝑥 + 6| = -3
8. |2𝑥 − 3| = 0
Answers
1. x = 6, or -6
5. x = -2 or 2
2. x = 0
3. No solution
4. x = 3 or 5
6. x = 7 or -13/3
7. No solution
8. x = 3/2
Absolute value inequalities
You can now combine your knowledge of inequalities with your understanding of absolute
value. Study these examples to understand how the two concepts work when used in the same
equation.
Example 1:
|𝑥| < 4
-4 < x < 4
This inequality indicates that x is a number which lies less than 4 units from zero on the
number line. That means that x can lie anywhere between -4 and 4.
Example 2:
|𝑥| > 4
x < -4
or x > 4
This inequality states that x lies more than 4 units from zero on the number line.
Therefore, x must be to the left of -4 or to the right of 4.
Example 3:
|6𝑥 + 7| < 5
-5 < 6x + 7 < 5
-12 < 6x < -2
-2 < x < -1/3
[The < (less than) symbol indicates that 6x + 7 is greater than -5 and less than 5.]
Example 4:
|2𝑥 − 9| > 4
2x – 9 < -4
2x < 5
5
x<2
Multiplication symbols: *
or
2x – 9 > 4
2x > 13
13
x> 2
● or x
November 2012 Document1
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[The > (greater than) symbol indicates a distance of more than 4 units, therefore it can be
interpreted as meaning or.]
Now it’s your turn. Here are some practice problems for you.
1. |𝑥| < 3
2. |𝑦| > 12
3. |𝑥 + 4| < 9
4. |𝑥 − 2| > 6
5. |5𝑥 + 2| < 13
6. |9 − 4𝑥| > 14
7. |−5 − 7𝑥| < 30
8. |2 − 9𝑥| > 17
9. |𝑚 + 5| + 9 < 16
10. 7 - |3 − 2𝑥| > 5
Answers
1. -3<x<3
2. y < -12 or y > 12
5. -3 < x < 11/5 6. x < -5/4 or x > 23/4
9. -12 < m < 2
10. ½ < x < 5/2
3. -13 < x < 5
7. -5 < x < 25/7
4. x < -4 or x > 8
8. x < -5/3 or x > 19/9
Location and identification of points on the plane
On a number line every number can be represented by a point on the line. When you graph
equations in two variables, it is necessary to use two numbers to designate each point. Points are
graphed on a coordinate plane which is made up of two perpendicular lines. The pairs of points
are referred to as ordered pairs because it is important what order the pair of numbers is in. The
first number refers to the distance and direction from zero along the horizontal axis; the second
number refers to the vertical distance and direction from zero.
Multiplication symbols: *
● or x
November 2012 Document1
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Notice that the points (3,2) and (2,3) are not the same point; similarly, the points (-4,5) and (5,-4)
are not the same point. This is because the first coordinate indicates horizontal distance and the
second indicates vertical distance.
Distance formula
The formula 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 is used to find the distance between any two points.
Example 1:
Find the distance between (4, -3) and (-5, 4).
First, decide which is point 1 and which is point 2.
x1 = 4, y1 = -3 and x2 = -5, y2 = 4
(note: it does not matter which point you call point 1 and which you call point 2.
The answer will come out the same.)
𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑑 = √(−5 − 4)2 + (4 − (−3))2
𝑑 = √(−9)2 + (7)2
𝑑 = √81 + 49
𝑑 = √130 ≈ 11.402
Try these problems.
Find the distance between the two points.
1. (2, 6) and (-4, -2)
2. (-2, 1) and (4, 2)
Multiplication symbols: *
● or x
November 2012 Document1
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3. (-3, 1) and (6, -7)
4. (10, -7) and (8, -3)
Answers
1. 10
2. √37 ≈ 6.083
3. √145 ≈ 12.042
4. √20 ≈ 4.472
Exponential and logarithmic equations
Equations which have variables in exponents are called exponential equations. An example of
such an equation is 23x = 64. Often these types of equations can be solved by writing both sides
using the same base. If the bases are equal, then the exponents must also be equal.
Example 1:
23x = 64
23x = 26
Since the bases are the same, the exponents are equal.
3x = 6
x=2
Example 2:
54x+7 = 125
54x+7 = 53
4x + 7 = 3
4x = -4
x = -1
Sometimes it’s impractical – or seemingly impossible – to write both sides of an equation using
the same base. In that case you can solve the equation by taking either the common logarithm or
the natural logarithm of both sides.
Example 3:
5x = 12
𝑙𝑜𝑔 5x = log 12
x log 5 = log 12
𝑙𝑜𝑔12
1.0792
x = 𝑙𝑜𝑔5 ≈ 0.6990 ≈ 1.544
In a similar way logarithmic equations can sometimes be changed to exponential equations to
make it easier to solve them.
Example 4:
log2 x = 3
x = 23
x=8
Solve for x using any method that seems appropriate.
1. 2x = 8
Multiplication symbols: *
● or x
November 2012 Document1
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2. 32x = 9
3. 42x-3 = 64
4. 7x = 20
5. 2x = 55
6. log5x = 2
7. log3(5x + 7) = 2
8. log2 x = -5
Answers
1. 3
2. 1
3. 3
4. Approx. 1.54
5. Approx. 5.78
6. 25
7. 2/5 8. 1/32
Factorials
Products such as 8●7●6●5●4●3●2●1 are often used in higher mathematics and especially in
statistics. Because it is seen so frequently, this type of product has its own symbol, “!”. The
product 8●7●6●5●4●3●2●1 can be written 8!, which is read “8” factorial.” 3! = 3●2●1 = 6.
You may also need to perform some arithmetic operations involving factorials. To make it
possible for certain statistical formulas to work properly, 0! is defined as equal to 1.
Example 1:
Example 2:
5! = 5●4●3●2●1 = 120
10!
5!
=
10∗9∗8∗7∗6∗5∗4∗3∗2∗1
5∗4∗3∗2∗1
=
3,628,800
120
= 30,240
Evaluate the following factorials.
1. 7!
2.
3.
9!
5!
8!
4!
4. 1!
5. 0!
Multiplication symbols: *
● or x
November 2012 Document1
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6.
10!
7!∗3!
7. (8-5)!
8.
7!
(7−2)!
Answers
1. 5,040
2. 3,024
3. 1, 680
4. 1
5. 1
6. 120
7. 6
8. 42
Evaluation of functions
A function is a specific, carefully defined relationship between two sets of items. For the
relationship to be considered a function, it must be a one-to-one relationship, that is each item in
the first set must correspond to exactly one item in the second set. Let’s look at some examples
that do not involve numbers.
Example 1: Each state in the United States has one and only one capital, therefore it is a oneto-one relationship. It could be considered a function.
Example 2: If you want to buy a bicycle, each bike has exactly one price. This is a one-to-one
correspondence and could be considered a function.
Example 3: If you drive your car at 50 mph for 3 hours, you will travel 150 miles. If you
drive at a rate of 60 mph for 4 hours, you will travel 240 miles. Each rates and time correlates to
a distance. Distance is said to be a function of rate and time.
The same principle applies to many algebraic relationships. Consider this set of ordered pairs:
{(-2, 5), (5, 7), (0, 1), (4, -2)}. This relations is a function because none of the pairs have the
same x coordinate. Looking at the pairs in a different form may help clarify this idea.
x
-2
5
0
4
→
→
→
→
→
y
5
7
1
-2
Now consider this set of ordered pairs: {(9, -5), (9, 5), (2, 4)}. This is a not a function because
two of the ordered pairs have the same first coordinate and different second coordinates.
Multiplication symbols: *
x
9
→
→
2
→
y
-5
5
4
● or x
November 2012 Document1
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Functions used in mathematics are often given as equations. They are evaluated in the same way
that other equations are; only the notation differs.
In a previous section we evaluated polynomials by substituting numbers for variables as follows:
For x = 3, evaluate y = x2 – 5.
y = 32 – 5
y=9–5
y=4
Since this relationship is a function, you could also write f(x) = x2 – 5. This is read “f of x equals
x squared minus five.” (Note: The use of the letter f is arbitrary. You could also use other
letters such as g(x), h(x), d(f). The meaning is the same regardless of the letters used.) If you
were being asked to evaluate the expression for x = 3, it would appear something like this:
Find f(3) when f(x) = x2 – 5.
f(x) = x2 – 5
f(3) = 32 – 5
f(3) = 9 – 5
f(3) = 4
Functions can also be evaluated by replacing the x with another letter:
Find h(t) when h(x) = 7x + 4
h(x) = 7x + 4
h(f) = 7t + 4
Evaluate these functions for the values of x given.
1. g(x) = 3x2 – 2x +1
a) g(0)
b) g(-1)
c) g(3)
2. f(x) = 5x2 + 4x
a) f(0)
b) f(3)
c) f(-1)
𝑥−4
3. f(x) = 𝑥+3
a) f(5)
b) f(-3)
c) f(4)
Multiplication symbols: *
● or x
November 2012 Document1
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Answers
1. a) 1
b) 6
c) 22
2. a)0
b) 57
c) 1
3. a) 1/8
b) undefined
c) 0
Functions can also be combined mathematically, this is, they can be added, subtracted,
multiplied or divided.
Example 1:
f(x) = x + 2 and g(x) = x2 + 1
(f+g)(x) + (x + 2) + (x2 + 1)
= x2 + x + 3
(f – g)(x) = (x + 2) – (x2 + 1)
= -x2 + x + 1
(g – f)(x) = (x2 + 1) – (x + 2)
= x2 – x – 1
f(x) ● g(x) = (x + 2) ● (x2 + 1)
= x3 + 2x2 + x + 2
𝑓(𝑥)
𝑥+2
𝑔(𝑥)
= 𝑥 2 +1
𝑔(𝑥)
𝑥 2 +1
=
𝑓(𝑥)
𝑥+2
for x ≠ -2 (division by 0 is undefined)
Okay, it’s time for you to try it. Find (f+g)(x), (f – g)(x), (g – f)(x), fg(x), ff(x), f/g(x), and g/f(x)
for each pair of functions.
1. f(x) = x2 – 1
2. f(x) = x 2
3. f(x) = x2 – 4
g(x) = 2x + 5
g(x) = √𝑥
g(x) = x2 + 2
Answers
1. f + g = x2 + 2x + 4
Multiplication symbols: *
● or x
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f – g = x2 – 2x – 6
g – f = -x2 + 2x + 6
f – g = 2x3 + 5x2 – 2x – 5
f ● f = x4 – 2x2 + 1
𝑥 2 −1
𝑓
5
= 2𝑥+5 , x≠-2
𝑔
𝑔
𝑓
2𝑥+5
= 𝑥 2 −1, x≠1 or -1
2. f + g = x2 + √𝑥
f – g = x 2 - √𝑥
g – f = √𝑥 – x2
f – g = x 2 ● √𝑥
f ● f = x4
𝑓
𝑥2
=
𝑔
√𝑥
𝑔
√𝑥
𝑓
, x>0
= 𝑥 2 , x±0
3. f + g = 2x2 – 2
f – g = -6
g–f=6
f ● g = x4 – 2x2 – 8
f ● f = x4 – 8x2 + 16
𝑓
𝑥 2 −4
𝑔
𝑥 2 +2
= 𝑥 2 +2
𝑔
= 𝑥 2 −4, x±2 or -2
𝑓
Composition of functions
If one function’s output (answer) depends on the output or answer of another function, they are
called composite functions. One real life example of functions depending on each other is the
calculation of your state income tax: the amount of state income tax you must pay depends on
your adjusted gross income reported on your federal tax return, which in turn depends on your
annual earnings. These functions are composites.
Multiplication symbols: *
● or x
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The symbol for composition of functions an open circle: f ○ g. This is read “f composed with
g,” “the composition of f and g,” and “f circle g.”
To compose one function with another, the variable in the first function can be replaced by the
entire second function.
Example 1:
Given that f(x) = 2x – 5
and g(x) and x2 – 3x + 8
(f ○ g)(x) = 2(x2 – 3x + 8) – 5 = 2x2 – 6x + 11
And
(g ○ f)(x) = (2x – 5)2 – 3(2x – 5) + 8 = 4x2 – 26x + 48
You can also compose two functions by replacing the variable in one function with the answer to
the other function.
Example 2:
given the functions in example 1, find (f ○ g)(7) and (g ○ f)(7)
For (f ○ g)(7)
g(7) = 72 – 3(7) + 8 = 36
then (f ○ g)(7) = 2 ●36 – 5 = 67
For (g ○f)(7):
(f)(7) = 2 ● 7 – 5 = 9
Then (g ○ f)(7) = 92 – 2 ● 9 + 8 = 62
Composition of functions can be tricky. Remember, you are replacing the variable in one
function with the other function. Practice with these functions. Find both f ○ g and g ○ f.
1. f(x) = √𝑥
and
g(x) = x – 3
2. f(x) = x5
and
g(x) = 2x – 3
Answers
1. (f ○ g)(x) = √𝑥 − 3
(g ○ f)(x) = √𝑥 -3
2. (f ○ g)(x) = (2x – 3)5
(g ○ f)(x) = 2x5 – 3
Now repeat the composition of functions for x = 12.
Answers
Multiplication symbols: *
● or x
November 2012 Document1
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1. (f ○ g)(12) = 3
(g ○ f)(12) = √12 – 3 or 2√3 – 3
2. (f ○ g)(12) = 215
(g ○ f)(12) = 2(125) – 3
Multiplication symbols: *
● or x
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