§ 5.7 polynomial equations and their applications

§ 5.7

Polynomial Equations and Their Applications

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.



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


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



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 ?

?

?

?


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



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




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



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



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



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



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 ? ?



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



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



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




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


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)


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


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


Polynomial Equations in ApplicationCONTINUECONTINUE

DD

0

2

4

6

8

10

0 0.5 1 1.5 2

Time (seconds)

Heig

ht (f

eet)


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


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



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



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.

§ 5.7 polynomial equations and their applications

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