§ 5.7 Polynomial Equations and Their Applications.
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Transcript of § 5.7 Polynomial Equations and Their Applications.
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§ 5.7
Polynomial Equations and Their Applications
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Blitzer, Intermediate Algebra, 4e – Slide #91
Solving Polynomial Equations
Definition of a Quadratic EquationA quadratic equation in x is an equation that can be written in the standard form
where a, b, and c are real numbers, with . A quadratic equation in x is also called a second-degree polynomial equation in x.
02 cbxax0a
The Zero-Product RuleIf the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.
If AB = 0, then A = 0 or B = 0.
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Blitzer, Intermediate Algebra, 4e – Slide #92
Solving Polynomial Equations
Solving a Quadratic Equation by Factoring1)If necessary, rewrite the equation in the standard form , moving all terms to one side, thereby obtaining zero on the other side.2) Factor completely.3) Apply the zero-product principle, setting each factor containing a variable equal to zero.4) Solve the equations in step 3.5) Check the solutions in the original equation.
02 cbxax
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Blitzer, Intermediate Algebra, 4e – Slide #93
Solving Polynomial EquationsEXAMPLEEXAMPLE
SOLUTIONSOLUTION
.xx 4542 Solve:
1) Move all terms to one side and obtain zero on the other side. Subtract 45 from both sides and write the equation in standard form.
45454542 xx04542 xx
Subtract 45 from both sidesSimplify
2) Factor.
059 xx Factor
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Blitzer, Intermediate Algebra, 4e – Slide #94
Solving Polynomial Equations
4542 xx
3) & 4) Set each factor equal to zero and solve the resulting equations.
09 x or
CONTINUECONTINUEDD
05 x
9x 5x
5) Check the solutions in the original equation.Check 9: Check -5:
4542 xx
45949 2 45545 2
459481 455425 ?
?
?
?
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Blitzer, Intermediate Algebra, 4e – Slide #95
Solving Polynomial EquationsCONTINUECONTINUE
DD Check 9: Check -5:
453681 452025
4545 4545
The solutions are 9 and -5. The solution set is {9,-5}.
? ?
-60
-40
-20
0
20
40
60
80
-10 -5 0 5 10 15
The graph of
lies to the right.
4542 xxy
truetrue
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Blitzer, Intermediate Algebra, 4e – Slide #96
Solving Polynomial EquationsEXAMPLEEXAMPLE
SOLUTIONSOLUTION
x.x 42 Solve:
1) Move all terms to one side and obtain zero on the other side. Subtract 4x from both sides and write the equation in standard form. NOTE: DO NOT DIVIDE BOTH SIDES BY x. WE WOULD LOSE A POTENTIAL SOLUTION!!!
xxxx 4442
042 xxSubtract 4x from both sidesSimplify
2) Factor.
04 xx Factor
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Blitzer, Intermediate Algebra, 4e – Slide #97
Solving Polynomial Equations
3) & 4) Set each factor equal to zero and solve the resulting equations.
0x or
CONTINUECONTINUEDD
04 x
4x
5) Check the solutions in the original equation.Check 0: Check 4:
040 2 444 2
00 1616
? ?
xx 42 xx 42
truetrue
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Blitzer, Intermediate Algebra, 4e – Slide #98
Solving Polynomial EquationsCONTINUECONTINUE
DDThe solutions are 0 and 4. The solution set is {0,4}.
The graph of
lies to the right.
xxy 42
-10
0
10
20
30
40
50
60
70
80
-10 -5 0 5 10 15
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Blitzer, Intermediate Algebra, 4e – Slide #99
Solving Polynomial EquationsEXAMPLEEXAMPLE
SOLUTIONSOLUTION
.xx 1441 Solve:
Be careful! Although the left side of the original equation is factored, we cannot use the zero-product principle since the right side of the equation is NOT ZERO!!
1) Move all terms to one side and obtain zero on the other side. Subtract 14 from both sides and write the equation in standard form.
Simplify 14141441 xx
01441 xx
Subtract 14 from both sides
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Blitzer, Intermediate Algebra, 4e – Slide #100
Solving Polynomial EquationsCONTINUECONTINUE
DD2) Factor. Before we can factor the equation, we must simplify it first.
01441 xx014442 xxx
01832 xx
FOILSimplify
Now we can factor the polynomial equation.
063 xx
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Blitzer, Intermediate Algebra, 4e – Slide #101
Solving Polynomial EquationsCONTINUECONTINUE
DD3) & 4) Set each factor equal to zero and solve the resulting equations.
03 x or 06 x
6x3x
5) Check the solutions in the original equation.Check 3: Check -6:
? ?
1441 xx 1441 xx
144313 144616
1472 1427 ? ?
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Blitzer, Intermediate Algebra, 4e – Slide #102
Solving Polynomial EquationsCONTINUECONTINUE
DDCheck 3: Check -6:
true1414 true1414
The solutions are 3 and -6. The solution set is {3,-6}.
The graph of
lies to the right.
1832 xxy
-40
-20
0
20
40
60
80
100
120
-15 -10 -5 0 5 10 15
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Blitzer, Intermediate Algebra, 4e – Slide #103
Solving Polynomial EquationsEXAMPLEEXAMPLE
SOLUTIONSOLUTION
.xxx 022 23 Solve by factoring:
1) Move all terms to one side and obtain zero on the other side. This is already done.
+
2) Factor. Use factoring by grouping. Group terms that have a common factor.
23 2xx 2 x 0
Common factor is
Common factor is -1.2x
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Blitzer, Intermediate Algebra, 4e – Slide #104
Solving Polynomial Equations
0222 xxx
CONTINUECONTINUEDD
012 2 xx
0112 xxx
Factor from the first two terms and -1 from the last two terms
2x
Factor out the common binomial, x – 2, from each term
Factor completely by factoring as the difference of two squares
12 x
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Blitzer, Intermediate Algebra, 4e – Slide #105
Solving Polynomial Equations
3) & 4) Set each factor equal to zero and solve the resulting equations.
CONTINUECONTINUEDD
02 x 01x 01xor or2x 1x 1x
5) Check the solutions in the original equation. Check the three solutions 2, -1, and 1, by substituting them into the original equation. Can you verify that the solutions are 2, -1, and 1?
The graph of
lies to the right.
22 23 xxxy
-150
-100
-50
0
50
100
150
-6 -4 -2 0 2 4 6
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Blitzer, Intermediate Algebra, 4e – Slide #106
Polynomial Equations in ApplicationEXAMPLEEXAMPLE
A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The function describes the height of the gymnast’s feet above the ground, s (t), in feet, t seconds after dismounting. The graph of the function is shown below.
8816 2 ttts
0
2
4
6
8
10
0 0.5 1 1.5 2
Time (seconds)
Heig
ht (f
eet)
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Blitzer, Intermediate Algebra, 4e – Slide #107
Polynomial Equations in Application
SOLUTIONSOLUTION
When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph.
We note that the graph of the equation passes through the line y = 8 twice. Once when x = 0 and once when x = 0.5. This can be verified by determining when y = s (t) = 8. That is,
CONTINUECONTINUEDD
8816 2 ttts88168 2 tt
tt 8160 2
1280 tt
Original equationReplace s (t) with 8Subtract 8 from both sidesFactor
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Blitzer, Intermediate Algebra, 4e – Slide #108
Polynomial Equations in Application
Now we set each factor equal to zero.CONTINUECONTINUE
DD
08 t
We have just verified the information we deduced from the graph. That is, the gymnast will indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5 seconds. These solutions are identified by the dots on the graph on the next page.
012 tor0t 12 t
21
t
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Blitzer, Intermediate Algebra, 4e – Slide #109
Polynomial Equations in ApplicationCONTINUECONTINUE
DD
0
2
4
6
8
10
0 0.5 1 1.5 2
Time (seconds)
Heig
ht (f
eet)
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Blitzer, Intermediate Algebra, 4e – Slide #110
The Pythagorean Theorem
The Pythagorean TheoremThe sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.
If the legs have lengths a and b, and the hypotenuse has length c, then
.222 cba
A C
B
ca
b
Hypotenuse
Leg
Leg
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Blitzer, Intermediate Algebra, 4e – Slide #111
The Pythagorean TheoremEXAMPLEEXAMPLE
SOLUTIONSOLUTION
A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.
Let’s begin with a picture.
Tree
5 feet
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Blitzer, Intermediate Algebra, 4e – Slide #112
The Pythagorean Theorem
Since the wire is 1 foot longer than the height that it reaches on the tree, if we call the length of the wire (the quantity we wish to determine) x, then the height of the tree would be x -1.
Tree
5 feet
CONTINUECONTINUEDD
x-1
x
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Blitzer, Intermediate Algebra, 4e – Slide #113
The Pythagorean Theorem
We can now use the Pythagorean Theorem to solve for x, the length of the wire.
CONTINUECONTINUEDD
222 51 xx
22 2512 xxx 22 262 xxx
0262 xx226
x13
This is the equation arising from the Pythagorean TheoremSquare x – 1 and 5Add 1 and 25Subtract from both sides2xAdd 2x to both sidesDivide both sides by 2
Therefore, the solution is x = 13 feet.