Part II: The Answers
142
Answers
1–100
45.
no real solutions
Factoring this polynomial by trial and error gives you the following:
x
x
x
x
8
4
4
4
12
35 0
5
7 0
+
+ =
+
(
) +
(
) =
Setting the first factor equal to zero gives you x
4
+ 5 = 0 so that x
4
= –5, which has no real
solutions. Setting the second factor equal to zero gives you x
4
+ 7 = 0 so that x
4
= –7,
which also has no real solutions.
46.
x = –1, 1
Factor the polynomial repeatedly to get the following:
x
x
x
x
x
x
x
4
2
2
2
2
3
4 0
1
4
0
1
1
4
0
+
− =
−
(
) +
(
) =
−
+
+
(
) =
(
)(
)
Now set each factor equal to zero and solve for x. Setting the first factor equal to zero
gives you x – 1 = 0 so that x = 1. The second factor gives you x + 1 = 0 so that x = –1.
And the last factor gives you x
2
+ 4 = 0 so that x
2
= –4, which has no real solutions.
47.
x = –3, 3
Factor the polynomial repeatedly to get the following:
x
x
x
x
x
x
4
2
2
2
81 0
9
9 0
3
3
9 0
− =
−
(
) +
(
) =
−
+
+
(
) =
(
)(
)
Now set each factor equal to zero and solve for x. Setting the first factor equal to zero
gives you x – 3 = 0 so that x = 3. Setting the second factor equal to zero gives you x + 3 = 0
so that x = –3. The last factor gives you x
2
+ 9 = 0, or x
2
= –9, which has no real solutions.
48.
x = 1, 9
5
To solve an absolute value equation of the form a b
= , where b > 0, you must solve the two
equations a = b and a = –b. So for the equation 5
7 2
x − = , you have to solve 5x – 7 = 2:
5
7 2
5
9
9
5
x
x
x
− =
=
=
You also have to solve 5x – 7 = –2:
5
7
2
5
5
1
x
x
x
− = −
=
=
Therefore, the solutions are x = 9
5
and x = 1.