Chapter 1.6 Part 3 Equations and Inequalities.pdf
Transcript of Chapter 1.6 Part 3 Equations and Inequalities.pdf
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Fundamental Theorem of Algebra
Every polynomia l equat ion 0
with complex coefficients has a t least
one root .
P x
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Theorem
2
3 2
Every polynomia l of degree can be
expressed as product of linea r factors.
3 2 degree: 2
= 2 1
6 11 6 degree: 3
= 1 2 3
n
n
x x
x x
x x x
x x x
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Theorem
1 21 2
1 2
1 2
Every polynomia l equa t ion 0
of degree has a t most dist inct roots.
In genera l, a polynomia l equa t ion can be
writ ten as
0
, ,..., and a re the dist inct roots and
...
mk k k
m
m
P x
n n
P x a x r x r x r
r r r
k k
m
k n
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1 21 2
if
1, is a simple root .
2, is a double root .
, is a root of mult iplicity .
mk k k
m
i i
i i
i i
P x a x r x r x r
k r
k r
k m r m
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Example 1.6.12
2 4
Determine the roots of
1 3 5 0
dist inct roots: 1, 3, 5
1 is a simple root .
3 is a double root .
5 is a root of mult iplicity 4.
P x x x x
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Theorem
A polynomia l equat ion 0 of degree
has exact ly roots, a root of mult iplicity
being counted as roots.
P x
n n
k k
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Example 1.6.13
2
Form an equa t ion which has
1 as a double root
2 and 4 as simple roots
and no others.
1 2 4 0x x x
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Theorem
Let 0 be a polynomia l equa t ion with
rea l coefficients. Then if 0 has a
complex root , then it s conjuga te is a lso a
root of the equa t ion.
P x
P x
a bi a bi
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Example 1.6.14
Form an equa t ion with in tegra l
coefficients and with lowest possible
degree having 1 2 and 2 3 as
roots.
12 3 2 3 0
2
i
x x i x i
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2 2
2
2
12 3 2 3 0
2
2 1 2 3 2 3 0
2 1 2 3 2 3 0
2 1 2 3 0
2 1 4 4 9 1 0
2 1 4 13 0
x x i x i
x x i x i
x x i x i
x x i
x x x
x x x
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Theorem
The roots of 0 a re precisely
the addit ive inverses of the roots of
0.
2 is a root of 0
2 is a root of 0
P x
P x
P x
P x
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Example 1.6.15
5 3 2
5 3 2
5 3 2
5 3 2
Obta in an equa t ion whose roots a re
the nega t ives of the roots of
2 3 4 2 0
2 3 4 2 0
2 3 4 2 0
2 3 4 2 0
x x x x
P x x x x x
P x x x x x
x x x x
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Variation of Signs
descending powers
varia t ion of sign
If the terms of a re a rranged in
of , we say tha t
a occurs when two
successive terms have different signs.
P x
x
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Example 1.6.16
5 4 2
5 3 2
Determine the number of va r ia t ion of
signs for each polynomia l.
1. 2 3 4
var ia t ion of signs: 3
2. 2 3 4 2
var ia t ion of signs: 4
x x x x
x x x x
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Descartes Rule of Signs
The of the
polynomia l equa t ion 0 with
rea l coefficients is
number of posit ive roots
number of var ia t
equa l to the
in
or less than tha t by
ion of signs
an even number.
P x
P x
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Descartes Rule of Signs
The of 0
is
number of negat ive roots
number of posit ive roots the of 0.
P x
P x
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Example 1.6.17
7 4 3
7 4 3
Determine the possible number of posit ive,
nega t ive, and complex roots of
2 4 2 5 0
2 4 2 5 0
posit ive roots: 2 or 0
nega t ive roots: 3 or 1
complex roots: 6,4 or 2
P x x x x x
P x x x x x
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Rational Root Theorem
2
0 1 2
0
Consider
... 0, 0
with in tegra l coefficients.
If is a root , where and a re
rela t ively pr ime in tegers, then is a
factor of and is a factor of .
n
n n
n
a a x a x a x a
pp q
q
p
a q a
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Example 1.6.18
3 2
3 2
3 2
Solve 2 3 7 3 0
: 1, 3 : 1, 2
1 3: 1, 3, ,
2 2
2 3 7 3 0
2 3 7 3 0
posit ive roots: 3 or 1 nega t ive roots: 0
x x x
p q
p
q
x x x
x x x
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3 2
3 2
2 3 7 3 0
1 3: 1,3, ,
2 2
2 3 7 3
1 2 3 7 3
2 1 6
2 1 6
1 is not a root
3
x x x
p
q
x x x
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3 2
2
2 3 7 3
12 3 7 3
2
1 1 3
2 2 6
1 1 is a root and is a factor .
2 2
12 2 6 0
2
0
x x x
x
x x x
-
2
2
2
12 2 6 0
2
12 3 0
2
10 3 0
2
1 1 11
2 2
1 1 11 1 11, ,
2 2 2
x x x
x x x
x x x
ix x
i iS S
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Example 1.6.19
4 3 2
4 3 2
4 3 2
Solve 8 14 13 6 0
: 1, 2, 3, 6 : 1
: 1, 2, 3, 6
8 14 13 6 0
8 14 13 6 0
posit ive roots: 0
nega t ive roots: 4 or 2 or 0
x x x x
p q
pq
x x x x
x x x x
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4 3 2
4 3 2
3 2
8 14 13 6 0
: 1, 2, 3, 6
8 14 13 6
1 1 8 14 13 6
1 7 7 6
1 7 7 6 0
1 is a root , 1 is a factor .
1 7 7 6 0
x x x x
pq
x x x x
x
x x x x
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3 2
3 2
2
1 7 7 6 0
: 1, 2, 3, 6
7 7 6
6 1 7 7 6
6 6 6
1 1 1 0
6 is a root , 6 is a factor
1 6 1 0
x x x x
pq
x x x
x
x x x x
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2
2
1 6 1 0
1 0 6 0 1 0
1 6 1 1 1
1 1 4 1 1
2
1 3 1 3
2 2
1 3 1 31, 6, ,
2 2
x x x x
x x x x
x x a b c
x
i
i iS S
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Equations in Linear or Quadratic Forms
Some equa t ions can be converted to
linear or quadra t ic equa t ions.
equa t ions involving ra t iona l expression s
equa t ions involving radica ls
equa t ions in quadra t ic form
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Example 1.6.20
EQUATIONS & INEQUALITIES
Solution: The LCD of the RE is
Multiplying both sides by the LCD:
2
1 2 7
2 1 2x x x x
2 1 .x x
21 2 7
2 1 2 12 1 2
x x x xx x x x
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Checking the results shows that the LCD 0 for .
Therefore, the solution set is .
2 x
2 SS
1 2 2 7 x x 2 2 x x
21 2 7
2 1 2 12 1 2
x x x xx x x x
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Example 1.6.21
Solution:
The LCD of the fractions is .
Multiplying both sides by the LCD yields:
2
3 2 3
3 2 5 6x x x x
3 2 x x
3 2 2 3 3 x x3 6 2 6 3
3
x x
x
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Checking the results shows that the LCD = 0 for . Thus, 3 is NOT a solution, hence, there is NO SOLUTION to the equation.
Therefore, the solution set is .
3x
SS
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Example 1.6.22
3 3 1x x
2 2
3 3 1x x
square both sides of the
equat ion not the equat ion
23 9 6 1x x x
29 7 2 0x x 9 2 1 0x x
9 2 0 1 0x x
2
19
x x
3 1 3x x
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Checking:
2:
9x
1 :x
22 8
3 19 3
2 23
9 3
2 is an extraneous solut ion
9
2
1 3 1 3 3 1 3
1S S
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Example 1.6.23
2 5 1 2x x
2
2 5 1 4x x
22 5 2 2 3 5 1 4x x x x
23 4 2 2 3 5 4x x x
23 2 2 3 5x x x
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2 29 4 2 3 5x x x 23 2 2 3 5x x x
2 29 8 12 20x x x
2 12 20 0x x
10 2 0x x
10 0 2 0
10 2
x x
x x
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Checking:
10 :x 2 10 5 10 1 8 4
10 is an ext raneous root .
2 :x 2 2 5 2 1 4 2
2 is an ext raneous solut ion
SS
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Example 1.6.24
4 21 2x x 2
2 22 1 0x x
2
2 1 0x 2 1 0x
2 1x
x i
,SS i i
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Example 1.6.25
22 2
6 1 5 1 6 0x x
2
Let 1ux
26 5 6 0u u
2 3 3 2 0u u
3
2u
2
3u
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2 31
2x
4, 6
5S S
3
2u
2
3u
2 2
13x
2 4 3x x
5 4x
4
5x
3 6 2x x
6x