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Lesson 7.1 Exercises, pages 527–531
A
4. Which expressions are√rational expressions? Justify your choices.
x + 2
a
x + 3
a) 5
b) 4x
c) 7
√
3 3 b + 2b
x2 + 2x - 7
3x + 9
d) 2
e)
f)
2
x + 3
x - 2
16 - b
Parts a, c, and f are rational expressions because each expression is the
quotient of two polynomials. Parts b and e are not rational expressions
because they each contain the root of a variable; part d is not a rational
expression because it has a variable as an exponent.
5. Identify the non-permissible values of the variable for each rational
expression.
a)
2 - x2
x + 5
b)
x + 1
(x - 2)(x + 8)
(x ⴚ 2)(x ⴙ 8) ⴝ 0
x ⴚ 2 ⴝ 0 or x ⴙ 8 ⴝ 0
x ⴝ 2 or x ⴝ ⴚ8
So, x ⴝ 2 and x ⴝ ⴚ 8 are the
non-permissible values.
xⴙ5ⴝ0
x ⴝ ⴚ5
So, x ⴝ ⴚ5 is the
non-permissible value.
6. Determine whether the given value of x is a non-permissible value
for the rational expression. Explain how you know.
a)
3x + 9
;x = 6
(x - 5)(x + 6)
No, x ⴝ 6 does not result in
the denominator equal to 0.
©P
DO NOT COPY.
b)
5x
; x = -2
x - 4
2
Yes, x ⴝ ⴚ2 results in
the denominator equal to 0.
7.1 Equivalent Rational Expressions—Solutions
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7. Simplify each rational expression. State the non-permissible values
of the variables.
25mn
a) 5m
b)
The non-permissible value
is: m ⴝ 0
2x(x - 3)
x - 3
The non-permissible value
is: x ⴝ 3
2x(x ⴚ 3)
2x (x ⴚ 3)
ⴝ
xⴚ3
xⴚ3
5
25mn
25 m n
ⴝ
5m
5 m
ⴝ 5n, m ⴝ 0
c)
ⴝ 2x, x ⴝ 3
(x - 3)(x + 4)
(x + 4)(x + 6)
d)
3x
12x(x + 5)
The non-permissible values
are: x ⴝ ⴚ4 and x ⴝ ⴚ6
The non-permissible values
are: x ⴝ 0 and x ⴝ ⴚ5
(x ⴚ 3)(x ⴙ 4)
3x
12x(x ⴙ 5)
(x ⴙ 4)(x ⴙ 6)
ⴝ
(x ⴚ 3) (x ⴙ 4)
ⴝ
(x ⴙ 4) (x ⴙ 6)
4
ⴝ
xⴚ3
ⴝ
, x ⴝ ⴚ6, ⴚ 4
xⴙ6
3x
12x (x ⴙ 5)
1
, x ⴝ ⴚ5, 0
4(x ⴙ 5)
B
8. Determine the non-permissible values for each rational expression.
a)
x2 + 3
x - x - 20
b)
2
x2 ⴚ x ⴚ 20 ⴝ 0
(xⴚ5)(x ⴙ 4) ⴝ 0
So, x ⴝ 5 and x ⴝ ⴚ4
are the non-permissible
values.
c)
x2 ⴚ 36 ⴝ 0
(x ⴚ 6)(x ⴙ 6) ⴝ 0
So, x ⴝ ⴚ 6 and x ⴝ 6
are the non-permissible
values.
x(2x - 3)
d)
2
4x + 17x - 15
4x2 ⴙ 17x ⴚ 15
4x ⴙ 20x ⴚ 3x ⴚ 15
4x(x ⴙ 5) ⴚ 3(x ⴙ 5)
(4x ⴚ 3)(x ⴙ 5)
2
So, x ⴝ ⴚ 5 and x ⴝ
ⴝ
ⴝ
ⴝ
ⴝ
3
4
3x
x2 - 36
0
0
0
0
2x
12x + 2x
2
12x2 ⴙ 2x ⴝ 0
2x(6x ⴙ 1) ⴝ 0
1
6
So, x ⴝ ⴚ and x ⴝ 0
are the non-permissible
values.
are the non-permissible
values.
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7.1 Equivalent Rational Expressions—Solutions
DO NOT COPY.
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9. Which of these rational expressions are defined for all real values of x?
Explain how you know.
a)
2x3 + 3
6x
b)
x ⴝ 0 is a non-permissible
value, so the expression is
not defined for all values of x.
c)
3x2 + 2x - 1
x2 + 49
3x + 7
x2 - 9
x ⴝ 3 and x ⴝ ⴚ3 are nonpermissible values, so the
expression is not defined for
all values of x.
d)
x2 + 4
x3 + 1
x ⴝ ⴚ1 is a non-permissible
value, so the expression is
not defined for all values of x.
Since x2 » 0, x2 ⴙ 49>0
Since the denominator cannot
equal 0, the expression is
defined for all values of x.
10. Use multiplication and division to write two equivalent forms of
each rational expression.
a)
2x
2x + 4
b)
The expression has x ⴝ ⴚ2
as a non-permissible value.
(x ⴙ 1)
2x
2x
ⴝ
2x ⴙ 4
(2x ⴙ 4) (x ⴙ 1)
#
ⴝ
2x(x ⴙ 1)
(2x ⴙ 4)(x ⴙ 1)
This expression has an
additional non-permissible
value: x ⴝ ⴚ1
2x
2x
ⴝ
2x ⴙ 4
2 (x ⴙ 2)
x
ⴝ
xⴙ2
The equivalent expressions are:
2x(x ⴙ 1)
(2x ⴙ 4)(x ⴙ 1)
x
, x ⴝ ⴚ2
xⴙ2
, x ⴝ ⴚ2, ⴚ1
3(x + 5)
x(x + 8)(x + 5)
This expression has x ⴝ 0,
x ⴝ ⴚ8, and x ⴝ ⴚ5 as
non-permissible values.
3(x ⴙ 5)
x(x ⴙ 8)(x ⴙ 5)
ⴝ
ⴝ
3(x ⴙ 5)
3(x ⴙ 5)2
x(x ⴙ 8)(x ⴙ 5)2
3(x ⴙ 5)
x(x ⴙ 8)(x ⴙ 5)
ⴝ
ⴝ
3 (x ⴙ 5)
x(x ⴙ 8) (x ⴙ 5)
3
x(x ⴙ 8)
The equivalent expressions are:
3(x ⴙ 5)2
x(x ⴙ 8)(x ⴙ 5)2
and
©P
DO NOT COPY.
# (x ⴙ 5)
x(x ⴙ 8)(x ⴙ 5) (x ⴙ 5)
, x ⴝ ⴚ8,ⴚ5, 0
3
, x ⴝ ⴚ8,ⴚ5, 0
x(x ⴙ 8)
7.1 Equivalent Rational Expressions—Solutions
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11. a) Write each rational expression in simplest form.
i)
- p3q2
ii)
5p2q2
The non-permissible values
are: p ⴝ 0 and q ⴝ 0
ⴚ p3q2
5p2q2
ⴝ
ⴝ
iii)
ⴚp 3 1 q 2
5 p2 q 2
ⴚp
, p ⴝ 0, q ⴝ 0
5
2x3 + 4x2
6x2 - 24
ⴝ
iv)
2x2(x ⴙ 2)
6(x2 ⴚ 4)
2x (x ⴙ 2)
2
ⴝ
6(x ⴚ 2)(x ⴙ 2)
The non-permissible
values are:
x ⴝ ⴚ2 and x ⴝ 2
2 x2 (x ⴙ 2)
ⴝ
3
ⴝ
6 (x ⴚ 2) (x ⴙ 2)
x2
, x ⴝ ⴚ2, 2
3(x ⴚ 2)
4x - 9
x2 - 9
The non-permissible values are:
x ⴝ ⴚ3 and x ⴝ 3
The numerator and denominator
have no common factors, so the
expression is in simplest form:
4x ⴚ 9
, x ⴝ ⴚ3, 3
x2 ⴚ 9
36 - 9x2
x - 5x + 6
2
ⴝ
ⴝ
9(4 ⴚ x2)
(x ⴚ 2)(x ⴚ 3)
9(2 ⴚ x)(2 ⴙ x)
(x ⴚ 2)(x ⴚ 3)
The non-permissible
values are:
x ⴝ 2 and x ⴝ 3
ⴝ
ⴚ 9 (x ⴚ 2) (2 ⴙ x)
(x ⴚ 2) (x ⴚ 3)
ⴚ 9(2 ⴙ x)
, x ⴝ 2, 3
ⴝ
xⴚ3
b) Choose one expression from part a. Explain why the nonpermissible values of the given expression and its simplest
form are the same.
In part iii, the numerator and denominator of the expression were
divided by the common factor x ⴙ 2. This division is not possible
when x ⴝ ⴚ2. So, x ⴝ ⴚ 2 must be included in the non-permissible
values of the simplified expression.
12. Here is a student’s solution for simplifying a rational expression.
Identify the error in the solution. Write a correct solution.
3 (x - 4)
3x - 12
=
(x
+
5) (x - 4)
x + x - 20
3
=
, x Z -5
x + 5
2
The error in the solution is that x ⴝ 4 should be included as a nonpermissible value. Division by x ⴚ 4 is not possible when x ⴝ 4.
Correct solution:
3 (x ⴚ 4)
3x ⴚ 12
ⴝ
(x ⴙ 5) (x ⴚ 4)
x2 ⴙ x ⴚ 20
3
, x ⴝ ⴚ5, 4
ⴝ
(x ⴙ 5)
4
7.1 Equivalent Rational Expressions—Solutions
DO NOT COPY.
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12x2(x - 3)
12x2
13. A student claims that the expressions 15x and
are
15x(x - 3)
equivalent. Is the student correct? Explain.
No, the student is not correct. When x ⴝ 3, the first expression equals 2.4
and the second expression is not defined. So, the expressions are not
equivalent. The non-permissible values of both expressions must be stated.
14. Create a rational expression that has each set of non-permissible
values. Explain how you created each expression.
a) x Z 0, x Z 6
Choose a denominator so that when x ⴝ 0 or 6, the value of the
denominator is 0, and the value of the denominator is non-zero for
all other values of x; for example, x(x ⴚ 6)
Then write a polynomial in the numerator:
x2 ⴙ 1
x(x ⴚ 6)
b) x Z 4, x Z - 7
Choose a denominator so that when x ⴝ 4 or ⴚ7, the value of the
denominator is 0, and the value of the denominator is non-zero for
all other values of x; for example, (x ⴚ 4)(x ⴙ 7)
Then write a polynomial in the numerator:
ⴚ 2x2 ⴙ x
(x ⴚ 4)(x ⴙ 7)
C
15. Write each rational expression in simplest form.
State the non-permissible values of the variables.
a)
2x2 - 7xy + 6y2
x4 - 16y4
ⴝ
ⴝ
ⴝ
2x2 ⴚ 4xy ⴚ 3xy ⴙ 6y2
ⴝ
(x2 ⴚ 4y2)(x2 ⴙ 4y2)
2x(x ⴚ 2y) ⴚ 3y(x ⴚ 2y)
(x ⴚ 2y)(x ⴙ 2y)(x ⴙ 4y )
2
x4 - 5x2 + 4
x3 + 3x2 + 2x
2
(2x ⴚ 3y)(x ⴚ 2y)
(x ⴚ 2y)(x ⴙ 2y)(x2 ⴙ 4y2)
ⴝ
ⴝ
(2x ⴚ 3y) (x ⴚ 2y)
ⴝ
ⴝ
(x2 ⴚ 4)(x2 ⴚ 1)
x(x2 ⴙ 3x ⴙ 2)
(x ⴚ 2)(x ⴙ 2)(x ⴚ 1)(x ⴙ 1)
x(x ⴙ 1)(x ⴙ 2)
The non-permissible values are:
x ⴝ 0, x ⴝ ⴚ2, and x ⴝ ⴚ1
The non-permissible values are:
x ⴝ 2y and x ⴝ ⴚ2y
ⴝ
©P
b)
(x ⴚ 2) (x ⴙ 2) (x ⴚ 1) (x ⴙ 1)
x (x ⴙ 1) (x ⴙ 2)
(x ⴚ 2)(x ⴚ 1)
, x ⴝ ⴚ2, ⴚ1, 0
x
(x ⴚ 2y) (x ⴙ 2y)(x2 ⴙ 4y2)
2x ⴚ 3y
(x ⴙ 2y)(x2 ⴙ 4y2)
DO NOT COPY.
, x ⴝ 2y,ⴚ2y
7.1 Equivalent Rational Expressions—Solutions
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