Unformatted text preview: . How many real solutions does the equation
x 3 ax b 0
have in the case 27b 2 4a 3 0? What if 27b 2 4a 3 0?
3. Use Scientific Notebook to find the exact form of the solutions of the equation
125x 3 300x 2 195x 28 4 7 0
and show graphically that only one of these solutions is real. Exercises on Inequalities
1. Prove that if x and y are positive real numbers then their product xy is positive.We assume that x 0 and
y 0. Since 0 x and y 0 it follows from the order axiom for the real number system that
and the fact that 0y 0 allows us to deduce that 0 xy.
2. Prove that if x and y are negative real numbers then their product xy is positive.
If x and y are negative then x and y are positive and since
xy x y
the fact that xy is positive follows from Exercise 1.
3. Given real numbers a and b, prove that
|a | |b | |a b |. Since
a b a
it follows from the triangle inequality that 74 b |a | |b a b| |b | |a b| |a | |b | |a b |. ||a | |b || |a b |. |a | |b | |a b |. and therefore 4. Given real numbers a and b, prove that
From Exercise 3 we know that
|b | |a | |b a | |a b |
Since ||a | |b || is one of the numbers |a | |b | and |b | |a | the inequality
||a | |b || |a b |
follows at once.
5. In each of the following cases, find the numbers x for which the given inequality is true. Compare your
answers with the answers given by Scientific Notebook
a. |2x 3 | |6 x |.
Method 1: We separate the problem into three cases as illustrated in the figure: Case 1: When x 3
the inequality |2x 6
|6 x | says that 3
which tells us that x
Case 2: When 3 x
2 3| 2x 6 x 3.
6 the inequality |2x 3| |6 x | says that 3 6 x 2x
which tells us that x 3.
Case 3: When x 6 the inequality |2x 3 | |6 x | says that
2x 3 x 6
which tells us that x
3 (which is impossible). The set of numbers x for which the inequality
|2x 3 | |6 x | holds is therefore
3, 3 Þ 3 , 3 3, 3 .
Method 2: We look first at the equation
|2x 3 | |6 x |
which says that either 2x 3 6 x or 2x 3 x 6. The latter condition says that either x 3
or x 3 and so we separate the problem into the cases indicated by the next figure 3 3 This method leads us even more quickly to the solutions set 3, 3 . We omit the details. b. ||x | 5 | |x 6 |.
We provide one solution here. Another approach is suggested by the solution provided below
for part c. We begin by looking at the equation
||x | 5 | |x 6 |
which says that either |x | 5 x 6 or |x | 5 6 x. When x 0 these equations say that
either x 5 x 6 (which is impossible) or x 5 6 x, which tells us that x 11 .
2 75 When x 0 the equations say that either x 5 x 6 or x 5 6 x which are both
impossible. Therefore the only value of x at which the inequality ||x | 5 | |x 6 | can switch
from true to false or from false to true is 11 and by looking at a specimen value of x less than
and a specimen value greater then 11 we see that the inequality holds if and only if x 11 .
c. |2|x | 5| |4 |x 1...
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