Transcript Chapter Five 5.7
§ 5.7
Polynomial Equations and Their Applications
Solving Polynomial Equations
We have spent much time on learning how to factor polynomials.
Now we will look at one important use of factoring .
In this section, we will use factoring
to solve equations of degree 2
and higher. Up to this point, we have only looked at solving equations of degree one.
Blitzer,
Intermediate Algebra
, 5e – Slide #2 Section 5.7
Solving Polynomial Equations,
pp 368-369
Definition of a Quadratic Equation
A
quadratic equation
in
x
is an equation that can be written in the
standard form
ax
2
bx
c
0 where
a
,
b
, and
c
quadratic equation in
x
is also called a second-degree polynomial equation in
x
.
The Zero-Product Rule
If 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.
pp 368-369 Blitzer,
Intermediate Algebra
, 5e – Slide #3 Section 5.7
Solving Polynomial Equations,
p 369
Solving a Quadratic Equation by Factoring
1)If necessary, rewrite the equation in the standard form
ax
2
bx
c
0 , 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.
pp 369 Blitzer,
Intermediate Algebra
, 5e – Slide #4 Section 5.7
Solving Polynomial Equations
EXAMPLE (similar to example 1 on page 369) Solve:
x
2 4
x
45
.
SOLUTION
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.
x
2 4
x
45 45 45
x
2 4
x
45 0
2) Factor.
x
9
x
5 0 Subtract 45 from both sides Simplify Factor Blitzer,
Intermediate Algebra
, 5e – Slide #5 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting equations.
x
9 0
x
9 or
x
5 0
x
5
5) Check the solutions in the original equation.
Check 9: Check -5:
x
2 4
x
45
x
2 4
x
45 2 4 45 2 4 45 81 4 45 25 4 45 Blitzer,
Intermediate Algebra
, 5e – Slide #6 Section 5.7
Solving Polynomial Equations
CONTINUED
Check 9: Check -5:
81 36 45 45 45 true 25 20 45 45 45 true The solutions are 9 and -5. The solution set is {9,-5}.
The graph of lies to the right.
y
x
2 4
x
45 -10 -5 80 60 40 20 0 -20 0 -40 -60 5 10 Blitzer,
Intermediate Algebra
, 5e – Slide #7 Section 5.7
15
Solving Polynomial Equations
Check Point 1 Solve 2
x
2
9
x
5 2
x
2
9
x
5
0 2x 1
x
5 0 2x 1 0
x
5 0 x 1 2 x 5 1 2 , 5 P 370 Blitzer,
Intermediate Algebra
, 5e – Slide #8 Section 5.2
Solving Polynomial Equations
EXAMPLE (similar to example 2a on page 371 Solve:
x
2 4
x.
SOLUTION
1) Move all terms to one side and obtain zero on the other side.
Subtract 4
x
from both sides and write the equation in standard form.
NOTE: DO NOT DIVIDE BOTH SIDES BY x. WE WOULD LOSE A POTENTIAL SOLUTION!!!
x
2 4
x
4
x
4
x
Subtract 4
x
from both sides
x
2 4
x
0 Simplify
2) Factor.
x
x
4 0 Factor Blitzer,
Intermediate Algebra
, 5e – Slide #9 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting equations.
x
0 or
x
4 0
x
4
5) Check the solutions in the original equation.
Check 0: Check 4:
x
2 4
x x
2 4
x
2 4 2 4 0 0 true 16 16 true Blitzer,
Intermediate Algebra
, 5e – Slide #10 Section 5.7
Solving Polynomial Equations
CONTINUED The solutions are 0 and 4. The solution set is {0,4}.
The graph of
y
x
2 4
x
lies to the right.
-10 -5 40 30 20 10 80 70 60 50 0 -10 0 5 10 15 Blitzer,
Intermediate Algebra
, 5e – Slide #11 Section 5.7
Solving Polynomial Equations
EXAMPLE (similar to 2c on page 371 Solve:
x
1
x
4
14
.
SOLUTION
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.
x
x
1
x
1
x
4
4
14 14 14 14 0 Subtract 14 from both sides Simplify Blitzer,
Intermediate Algebra
, 5e – Slide #12 Section 5.7
Solving Polynomial Equations
CONTINUED
2) Factor.
Before we can factor the equation, we must simplify it first.
x
1
x
4 14 0
x
2 4
x
x
4 14 0 FOIL
x
2 3
x
18 0 Simplify Now we can factor the polynomial equation.
x
3
x
6 0 Blitzer,
Intermediate Algebra
, 5e – Slide #13 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting equations.
x
3 0
x
3 or
x
6 0
x
6
5) Check the solutions in the original equation.
Check 3: Check -6:
x
1
x
4
14
x
1
x
4
14
3 4
14
6 1
6 4
14 14 14 Blitzer,
Intermediate Algebra
, 5e – Slide #14 Section 5.7
Solving Polynomial Equations
CONTINUED
Check 3: Check -6:
14 14 true 14 14 true The solutions are 3 and -6. The solution set is {3,-6}.
The graph of
y
x
2 3
x
18 lies to the right.
-15 -10 -5 120 100 80 60 40 20 0 -20 0 -40 5 10 15 Blitzer,
Intermediate Algebra
, 5e – Slide #15 Section 5.7
Solving Polynomial Equations
EXAMPLE (similar to 2b on page 371) Solve by factoring:
x
3 2
x
2
x
2 0
.
SOLUTION
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.
x
3 2
x
2 +
x
2 0
Common factor is x
2
Common factor is -1.
Blitzer,
Intermediate Algebra
, 5e – Slide #16 Section 5.7
Solving Polynomial Equations
CONTINUED
x
2
x
2
x
2 0
x
2
x
2 1 0
x
2
x
1
x
1
0
x
2 Factor from the first two terms and -1 from the last two terms Factor out the common binomial,
x
– 2, from each term Factor completely by
x
2 factoring as the difference of two squares Blitzer,
Intermediate Algebra
, 5e – Slide #17 Section 5.7
Solving Polynomial Equations
CONTINUED
3) & 4) Set each factor equal to zero and solve the resulting equations.
x
2
x
0 2 or
x
1
x
0 1 or
x
1
x
0 1
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
y
x
3 2
x
2
x
2 lies to the right.
-6 -4 150 100 50 -2 0 -50 0 -100 -150 2 4 6 Blitzer,
Intermediate Algebra
, 5e – Slide #18 Section 5.7
Solving Polynomial Equations
Check Point 2a Solve 3
x
2 2
x
0 , 2 3 Check Point 2b Solve
x
2 7 10
x
18 Check Point 2c Solve
x
2
x
3 6 4 , 3 Blitzer,
Intermediate Algebra
, 5e – Slide #19 Section 5.7
P 372
Solving Polynomial Equations
Check Point 3 Solve 2
x
3 2
x
3 3
x
2 3
x
2 8
x
8
x
12 12 0
x
2x 2 2x 2x 3 2 3
x
2
x
4 3
x
2
x
2 3 0 0 0
2 ,
3 2 , 2 Blitzer,
Intermediate Algebra
, 5e – Slide #20 Section 5.7
P 370
Modeling Motion Application
EXAMPLE 4 See book p 373-4 (figure on p 374)
finds when the ball hits the ground building height is 384 similar to problem 65 on p. 378 similar to 98 on p. 388 Solve
s
16
t
2 32
t
384 0 16
t
2 32
t
384 Time to this height, substitute 0 for s(t) thrust
t
Beginning height
t
0 0 4 16 (
t
2 2
t
16
t
0
t
6 4
t
0 24 ) 6
4 t 4 , 6
6 seconds.
Factor out -16 factor The ball hits the ground after 6 Blitzer,
Intermediate Algebra
, 5e – Slide #21 Section 5.7
Modeling Motion Application
Check Point 4 on page 374
finds when the ball height is at 336 feet Solve
s
16
t
2 32
t
384 336 16
t
2 32
t
384 Time to this height, substitute 336 for s(t) 0 16
t
2 32
t
48
t
0
t
0 16 (
t
16 1
t
2
0
t
2
t
t
3 3 3 ) 0 1 t
1 , 3
3 Subtract 336 Factor out -16 factor The ball passes 336 after 3 seconds.
(3,336) is the point on the graph Blitzer,
Intermediate Algebra
, 5e – Slide #22 Section 5.7
Modeling Motion Application
Flight of a baseball
The height in feet of a baseball after t seconds is given by the following.
At what values of t is the height of the baseball 75 feet?
s
(
t
) 16
t
2 88
t
3 Time to this height, substitute 75 for s(t) thrust Beginning height 75 16
t
2 88
t
3 Time to this height, substitute 75 for h
Modeling Motion Application
Flight of a baseball, continued
75 16
t
2 88
t
3 75 16
t
2 88
t
3 0 16
t
2 88
t
3 75 16
t
2 88
t
72 0 8 ( 2
t
2 11
t
9 ) 0 8 2
t
9 0 Answer is 1 second going up and 4.5 seconds going down Multiply both sides by -1 Add 75 from both sides Combine like terms
t
t
1 1 0 Both valid
t
2
t
9 0 9 2 4 .
5
Modeling Motion Application
Model Rocket #98 on page 388
A model rocket is launched from the top of a cliff 144 feet above sea level.
The rocket misses the cliff and falls in the ocean. How long will it take for the rocket to hit the water?
s
(
t
) 16
t
2 128
t
144 Beginning height thrust 0 16
t
2 128
t
144 Time to this height, substitute 0 for s(t)
Modeling Motion Application
0 16
t
2 128
t
144
Model rocket #98 on page 388, continued
0 16 (
t
2 8
t
9 ) Factor out -16 16
t
9
t
1 0
t
t
9 9 0
t
t
1 1 0 Answer is 9 seconds
Modeling Motion Application
Model Rocket
A model rocket is launched from the ground.
How long will it take for the rocket to hit the ground?
Time to this height, substitute 0 for s(t) -16
t
= 0
t
= 0
s
(
t
) 16
t
2 48
t
0 16
t
2 48
t
0 16
t
(
t
3 ) Beginning height is zero t-3 = 0
t
= 3 The rocket strikes the ground after 3 seconds.
Modeling Motion Application
Car breaking, #99 on 388 (from extra credit)
How much distance do you need to bring your car to a complete stop?
d(x) is the stopping distance, in feet, for a care traveling at x miles per hour.
d
(
x
) 40 2
x
2
x
20 20
x x
800
x
2 20
x
It takes 40 feet to come to a complete stop, substitute 40 for d(x) 0
x
2 20
x
800 0 (
x
40 )(
x
20 ) X+40=0
t
= -40 X-20=0
t
= 20 The car was traveling 20 miles per hour.
Solving Polynomial Equations-Area
Do Check Point 5 on page 375 – see word problem diagram (also see Example 5 on page 374.)
Solve 12 2
x
16 2
x
320 192 24
x
32
x
4
x
2 320 4
x
2 56
x
192 320 4
x
2 4
x
56
x
16
x
128 2
0 0 factor
16 , 2
The path should be 2 feet wide.
P 375 Blitzer,
Intermediate Algebra
, 5e – Slide #29 Section 5.7
The Pythagorean Theorem,
p 375
The Pythagorean Theorem
The 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
length
c
, then
a
2
b
2
c
and
b
, and the hypotenuse has 2 .
B
Hypotenuse has length
c
The two legs have lengths of
a
and
b c a A b C
Blitzer,
Intermediate Algebra
, 5e – Slide #30 Section 5.7
The Pythagorean Theorem
EXAMPLE
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.
SOLUTION
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
x
Tree
x
-1 This is the equation arising from the Pythagorean Theorem
a
x
2
b
2 1 2
c
2 .
5 2
x
2 5 feet 5 feet Blitzer,
Intermediate Algebra
, 5e – Slide #31 Section 5.7
The Pythagorean Theorem
CONTINUED We can now use the
Pythagorean Theorem
to solve for
x
, the length of the wire.
x
1 2 5 2
x
2
x
2 2
x
1 25
x
2
x
2 2
x
2
x
26 26
x
2 0 26 2
x
13
x
This is the equation arising from the Pythagorean Theorem Square
x
– 1 and 5 Add 1 and 25
x
2 Subtract from both sides Add 2
x
to both sides Divide both sides by 2 Therefore, the solution is
x
= 13 feet.
Blitzer,
Intermediate Algebra
, 5e – Slide #32 Section 5.7
The Pythagorean Theorem
#82 on page 380
See picture on page 380 to solve for
x
, how far up the pole.
x
4 2
x
2 20 2
x
2 8
x
16
x
2 400 2
x
2 2
x
2 8
x
8
x
16 384 400 0 2 (
x
2 4
x
192 ) 2 (
x
12 )(
x
0 16 ) This is the equation arising from the Pythagorean Theorem Square
x +
4 and 20 Combine terms Subtract 400 from both sides GCF Factor
x
12 0
x
16 0
x
12
x
16 Therefore, the solution is
x
= 12 feet.
Blitzer,
Intermediate Algebra
, 5e – Slide #33 Section 5.7
Solving Polynomial Equations
Important to Note: A polynomial
equation
is the result of setting two polynomials
equal
other. to each The equation is in
standard form
if one side is 0 and the polynomial on the other side is in standard form, that is in descending powers of the variable.
A polynomial equation of
degree one is a linear is a quadratic
equation.
equation and of
degree two
Some polynomial equations can be solved by writing the equation in standard form, factoring, and then using the zero-product principle: If a product is 0, then at least one of its factors is equal to 0.
Blitzer,
Intermediate Algebra
, 5e – Slide #34 Section 5.7
DONE
Modeling Motion Application
h
16
t
2 88
t
3 Time to this height, substitute 75 for h thrust Beginning height 75 16
t
2 88
t
3 Time to this height, substitute 75 for h
h
16
t
2 132
t
0 thrust Beginning height (height of building) Time to this height, substitute 0 for h
Polynomial Equations in Application
EXAMPLE A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The
s
16
t
2 8
t
gymnast’s feet above the ground,
s
(
t
), in feet,
t
seconds after dismounting. The graph of the function is shown below.
10 8 6 4 2 0 0 0.5
1
Time (seconds)
1.5
2 Blitzer,
Intermediate Algebra
, 5e – Slide #37 Section 5.7
Polynomial Equations in Application
CONTINUED When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph.
SOLUTION 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,
s
16
t
2 8
t
8 8 16
t
2 8
t
8 Original equation Replace
s
(
t
) with 8 0 0 16
t
2 8
t
2
t
8
t
1
Subtract 8 from both sides Factor Blitzer,
Intermediate Algebra
, 5e – Slide #38 Section 5.7
Polynomial Equations in Application
CONTINUED Now we set each factor equal to zero.
8
t t
0 0 or 2
t
1 2
t
0 1
t
1 2 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.
Blitzer,
Intermediate Algebra
, 5e – Slide #39 Section 5.7
Polynomial Equations in Application
CONTINUED 10 8 6 4 2 0 0 2 0.5
1
Time (seconds)
1.5
Blitzer,
Intermediate Algebra
, 5e – Slide #40 Section 5.7