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Mathematics

Mathematics 0110A/B

Chris Brandl

Fall

Description

MATH 127 Fall 2012
Assignment 1 Solutions
1. Solve the following for x:
(a) 3jx + 2j = 2jx ▯ 1j ▯ 1
We have that x+2 = 0 when x = ▯2 and x▯1 = 0 when x = 1. Note that when
x = ▯2 the equation becomes 0 = 2j▯3j▯1 = 5 which is not possible. Similarly,
when x = 1 the equation becomes 3j3j = 0 ▯ 1 = ▯1 which again is not possible.
Thus, x = 1;▯2 are not solutions. We now consider the three intervals x < ▯2,
▯2 < x < 1, and x > 1.
For x < ▯2 we have that x + 2 < 0 and x ▯ 1 < 0 so our equation becomes
▯3(x + 2) = ▯2(x ▯ 1) ▯ 1
▯3x ▯ 6 = ▯2x + 1
▯3x + 2x = 1 + 6
▯x = 7
x = ▯7
This value of x lies within our interval and so, x = ▯7 is a solution.
Now consider the interval ▯2 < x < 1. For these values of x we get that x+2 > 0
and x ▯ 1 < 0 so our equation becomes
3(x + 2) = ▯2(x ▯ 1) ▯ 1
3x + 6 = ▯2x + 1
3x + 2x = 1 ▯ 6
5x = ▯5
x = ▯1
This value of x lies within our interval and so, x = ▯1 is a solution.
1 Lastly consider the interval x > 1. For these values of x we get that x + 2 > 0
and x ▯ 1 > 0 so our equation becomes
3(x + 2) = 2(x ▯ 1) ▯ 1
3x + 6 = 2x ▯ 3
3x ▯ 2x = ▯3 ▯ 6
x = ▯9
This value of x does not lie in our interval so there are no solutions where x > 1.
Hence, the two solutions to the equation are x = ▯7;▯1.
(b) jx ▯ 2j = x + 1 We have that x ▯ 2 ▯ 0 for x ▯ 2 and x ▯ 2 < 0 for x < 2.
For x ▯ 2 we have
x ▯ 2 = x + 1
x ▯ x = 1 + 2
0 = 3
which is impossible and so, there is no solution where x ▯ 2.
For x < 2 we have
▯(x ▯ 2) = x + 1
▯x + 2 = x + 1
▯x ▯ x = 1 ▯ 2
▯2x = ▯1
1
x =
2
2. Solve the following inequalities:
2 (a) 4 ▯ 4x ▯ 5 < 7
4 + 5 ▯ 4x < 7 + 5
9 ▯ 4x < 12
9
▯ x < 3
4
(b) j3 ▯ j < 2
x
4
▯2 < 3 ▯ x < 2
▯2 ▯ 3 < ▯ 4 < 2 ▯ 3
x
4
▯5 < ▯ < ▯1
x
5 1 1
> >
4 x 4
4 < x < 4
5
2 1
(c) > We have that 2x▯3 = 0 when x = 3 and x+6 = 0 when x = ▯6
2x ▯ 3 x + 6 2
so we look at the intervals x < ▯6, ▯6 < x < , and x > .
2 2
Consider when x < ▯6. For these values of x we have that both 2x▯3 and x+6
are negative so when we cross multiply we get that the inequality is
ipped twice
(stays the same) and so, we have
2(x + 6) > 2x ▯ 3
2x + 12 > 2x ▯ 4
2x ▯ 2x > ▯4 ▯ 12
0 > ▯16
which is always true. Thus, we get that for any x < ▯6 the original inequality
holds.
3
Now consider when ▯6 < x < 2. In this interval we get that x+6 is positive and
2x ▯ 3 is negative so when we cross multiply the inequality is
ipped and so, we
3 have
2(x + 6) < 2x ▯ 3
2x + 12 < 2x ▯ 4
2x ▯ 2x < ▯4 ▯ 12
0 < ▯16
which is impossible. Thus, there are no solutions in this interval.
3
Lastly consider when x >2. For these values of x we get that both x + 6 and
2x ▯ 3 are positive so when we cross multiply the inequality does not change and
so, we have 2(x + 6) > 2x ▯ 3 which we saw gives 0 > ▯16. This is always true
so for any x >3the inequality holds.
2
Hence, for any x >3or x < ▯6 the inequality holds.
2
(d) x + 2x ▯ 35 ▯ 0
Factoring the quadratic we get (x + 7)(x ▯ 5) ▯ 0. We see that when x = ▯7 or
x = 5 that the inequality is satis▯ed. We now need to consider the three intervals
x < ▯7, ▯7 < x < 5, and x > 5.
2
x + 7 x ▯ 5 x

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