Transcript Chapter Eight 8.1
Chapter 8 Quadratic Equations and Functions
§ 8.1
The Square Root Property and Completing the Square
Introduction
What we already know about quadratic equations… A quadratic equation can be written in the standard form:
ax
2
bx
c
0 ,
a
0 .
Some quadratic equations can be solved by factoring.
Some quadratic equations cannot be factored.
In this section, we look at a method for solving a quadratic equation that won’t factor. We look at a method called Completing the Square.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.1
The Square Root Property
The Square Root Property
If
u
is an algebraic expression and
d
is a nonzero real 2 number, then has exactly two solutions: If
u
2
d
, then
u
d
or
u
d
.
Equivalently, If
u
2
d
, then
u
d
.
This property says that when we take the square root of both sides of an equation, we get two roots. Don’t forget the P 563 Blitzer, Intermediate Algebra, 5e – Slide #4 Section 8.1
Using the Square Root Property
EXAMPLE Solve: 4
x
2 49 .
No bx SOLUTION To apply the square root property, we need a squared expression
x
2 by itself on one side of the equation. We can get by itself if we divide both sides by 4.
4
x
2 49
x
2 49 4 This is the given equation.
Divide both sides by 4.
x
x
49 4 49 4 7 2 3 .
5 Apply the square root property.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.1
Using the Square Root Property
CONTINUED
Check 3.5:
4 3 4 .
5
x
2 2 49 49 4 12 .
25 49 49 49 true
Check -3.5:
4 4 3 .
5
x
2 2 49 49 4 12 .
25 49 49 49 true The solutions are 3.5 and -3.5. The solution set is {3.5,-3.5}.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.1
Using the Square Root Property
Check point 1 page 564 Solve: 4
x
2 28 .
SOLUTION
x
2 Get by itself.
x
2 28 4
x x
28 4 4 7 4
x
7 No bx Divide both sides by 4.
Apply the square root property.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.1
Using the Square Root Property
CONTINUED
Check:
4 4
x
2 2 4 7 28 28 28 28 28 true
Check:
4 4
x
2 2 4 7 28 28 28 28 28 true Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.1
Using the Square Root Property
EXAMPLE Solve: SOLUTION 4
x
2 49 0 .
negative To solve by the square root property , we isolate the squared expression on one side of the equation.
4
x
2 49 0 4
x
2 49 This is the given equation.
Subtract 49 from both sides.
x
2 49 4 Divide both sides by 4.
x
49 4 Apply the square root property.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.1
Using the Square Root Property
CONTINUED
x
49 4 1
x
49 4
i
7 2
i
3 .
5
i
Check 3.5i:
4
x
2 49 0 4 2 49 0 4 4 12 .
25
i
2 49 12 .
25 0 49 0 0 0 true 1
i
4
Check -3.5i:
4 3 .
5
i x
2 2 49 49 0 0 4 4 12 .
25
i
12 .
25 2 49 49 0 0 0 0 true The solutions are 3.5
i
and -3.5
i
. The solution set is {3.5
i
,-3.5
i
}.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.1
Using the Square Root Property
Check point 2 Solve: SOLUTION 3
x
2 11 0 .
Isolate the squared expression on one side of the equation.
3
x
2 11
x
2 11 3
x
x
11 11 3 3 3 3 Add 11 from both sides.
Divide both sides by 3.
Apply the square root property.
Rationalize denominators.
Skipped in 7.5
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.1
Using the Square Root Property
CONTINUED
x
33 3 33 3 The solution set is Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.1
Using the Square Root Property
Check point 3 Solve: SOLUTION 4
x
2 9 0 .
Isolate the squared expression on one side of the equation.
4
x
2 9
x
2 9 4 Subtract 9 from both sides.
Divide both sides by 4.
x x
9 9 4 4 Apply the square root property.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.1
Using the Square Root Property
CONTINUED
x
9 4 9 4 1
x
3
i
2 3 The solution set is 2 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.1
Using the Square Root Property
EXAMPLE Solve by the square root property: 3
x
2 2 36 .
SOLUTION 3
x
2 2 36
x
2 2 12 This is the given equation.
Divide both sides by 3.
Isolate the squared expression on one side of the equation .
x
2
x
2 12 or
x
2 12 or
x
2 12 12 Apply the square root property.
Subtract 2 from both sides of each equation.
x
2
x
2 2 4 3 or
x
2 3 or
x
2 2 4 3 3 Rewrite radicands.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.1
Using the Square Root Property
CONTINUED 3 2 3
x
2 2 3 2 2 3 2 3 2 2 2 2 2 36 36 36 36 3 4 3 36 36 36 true 3
x
2 2 3 2 3 2 3 2 3 2 3 2 2 2 2 36 36 36 36 3 4 3 36 36 36 true 2 2 3 . Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.1
Completing the Square
Completing the Square
b
2 2 If is a binomial, then by adding , which is the square of half the coefficient of
x
, a perfect square trinomial will result. That is, 2
x
2
bx
b
2 2
b
2 .
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.1
Completing the Square
EXAMPLE What term should be added to the binomial so that it becomes a perfect square trinomial? Write and factor the trinomial.
x
2 10
x
SOLUTION
x
2 10
x x
2 10
x
25
x
5 2 Add 10 2 2 .
2 Add 25 to complete the square.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 8.1
Completing the Square
Check point 5 What term should be added to the binomial so that it becomes a perfect square trinomial?
a.
x
2 10
x x
2 10
x
25
x
5 2 b.
x
2 3
x
Add 10 2 2 .
2 Add 25 to complete the square.
2 3 2 9 4 .
x
2 3
x
9 4 3 2 2 c.
x
2 3 4
x
3 4 2 2 3 8 2 9 64
x
2 3 4
x
9 64 3 8 2 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.1
Completing the Square
Check point 6 on page 568 Solve by completing the square: SOLUTION
x
2 4
x
1 0 .
x
2 4
x
1
x
2 4
x
4 1 4
x
2 2 5
x
2
x
2 5 5 Add 1 to both sides.
Complete the square: 4 2 2 4 .
Factor and simplify.
The solution set is: 2 5 Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.1
Completing the Square
EXAMPLE (See example 7 on page 568 in textbook.) Solve by completing the square: 2
x
2 5
x
3 0 .
SOLUTION 2
x
2 5
x
3 0 This is the given equation.
x
2 5 2
x
3 2 0 Divide both sides by 2.
x
2 5 2
x
3 2
x
2 5 2
x
25 16 3 2 25 16 Add 3/2 to both sides.
Complete the square: Half of 5/2 is 5 4 , and 5 4 2 25 16 .
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.1
Completing the Square
CONTINUED 5 4 2 24 16 25 16 49 16
x
x
5 4 5 4 49 7 4 16 Factor and simplify.
Apply the square root property.
49 16 49 16 7 4
x
7 4 5 4 or
x
7 4 5 4 Split into two equations and subtract 5/4 from both sides of both equations.
x
2 4 or
x
12 4 Simplify.
1 The solutions are , and the solution set is 2 and 1 2 3 .
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 8.1
Completing the Square in Application
A Formula for Compound Interest
Suppose that an amount of money,
P
, is invested at rate
r
, compounded annually. In
t
years, the amount,
A
, or balance, in the account is given by the formula
A
P
1
r
t
.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.1
Completing the Square in Application
EXAMPLE: similar to #75 in homework
t
annual interest rate
r
.
In 2 years, an investment of $80,000 grows to $101,250.
SOLUTION We are given that
P
(the amount invested) = $80,000
t
(the time of the investment) = 2 years
A
(the amount, or balance, in the account) = $101,250.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.1
Completing the Square in Application
CONTINUED We are asked to find the annual interest rate,
r
. We substitute the three given values into the compound interest formula and solve for
r
.
A
101 , 250
P
1
r
80 , 000
t
1
r
2 101 , 250 1
r
2 80 , 000 81 64 1
r
2 Use the compound interest formula.
Substitute the given values.
Divide both sides by 80,000.
Simplify the fraction.
1
r
81 64 or 1
r
81 64 Apply the square root property.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 8.1
Completing the Square in Application
CONTINUED 1
r r
9 8 or 1 9 8 1 or
r r
9 8 9 8 1
r
1 8 or
r
17 8 81 64 Subtract 1 from both sides.
Simplify.
81 64 9 8 Because the interest rate cannot be negative, we reject -17/8. Thus, the annual interest rate is 1/8 = 0.125 = 12.5%.
t
$80,000 is invested for 2 years at 12.5% interest, compounded annually, the balance in the account is Blitzer, Intermediate Algebra, 5e – Slide #26 Section 8.1
Completing the Square in Application
CONTINUED
A
$ 80 , 000 1 0 .
125 2 $ 80 , 000 1 .
125 2 $ 101 , 250 .
Because this is precisely the amount given by the problem’s conditions, the annual interest rate is, indeed, 12.5% compounded annually.
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 8.1
Completing the Square in Application
Check Point 8 on page 570 - similar to #75 in homework
t
annual interest rate
r
.
You invested $3000 in an account whose interest is compounded annually. After 2 years, the amount, or balance, in the account is $4320. Find the annual interest rate.
SOLUTION 4320 3000 1
r
2 1 .
44 1
r
2 1
r
Divide both sides by 3,000.
Apply the square root property.
1 .
44 or 1
r
1 .
44 Blitzer, Intermediate Algebra, 5e – Slide #28 Section 8.1
Completing the Square in Application
CONTINUED 1
r
1 .
44 or 1
r
1 .
44 1
r
1 .
2 or 1
r
1 .
2
r
1 .
2 1 or
r
1 .
2 1 Subtract 1 from both sides.
r
0 .
2 or
r
2 .
2 Simplify.
Because the interest rate cannot be negative, we reject -2.2. Thus, the annual interest rate is 0.2 = 20%.
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 8.1
Square Root Property in Application
EXAMPLE (similar to #81 in homework)
s
(
t
) 16
t
2
s
(
t
) an object falls in t seconds. Use this function and the square root property to solve
Exercise 82
on page 573. A sky diver jumps from an airplane and falls for 3200 feet before opening a parachute. For how many seconds was the diver in a free fall? SOLUTION Note:
x
s
(
t
) 16
t
2 0
t
x
ie ( 16
t
2
bt
c
0 ) subtract x from both sides, multiply by 1
s
(
t
) 16
t
2 Blitzer, Intermediate Algebra, 5e – Slide #30 Section 8.1
Square Root Property in Application
CONTINUED
s
(
t
) 16
t
2 3200 16
t
2 200
t
2
t
200
t
100 2 Divide both sides by 16.
Square root of both sides Simplify
t
10 2 14 .
144 Disregard -14.1
Solution is 10 2 or 14 .
1 Blitzer, Intermediate Algebra, 5e – Slide #31 Section 8.1
Isosceles Right Triangles
Lengths Within Isosceles Right Triangles
The length of the hypotenuse of an isosceles right triangle is the length of a leg times 2 .
a a a 2 Blitzer, Intermediate Algebra, 5e – Slide #32 Section 8.1
Isosceles Right Triangles in Application
EXAMPLE: A supporting wire is to be attached to the top of a 70-foot antenna. If the wire must be anchored 70 feet from the base of the antenna, what length of wire is required?
SOLUTION Since the supporting wire, the antenna, and a line on the ground between the base of the antenna and the base of the supporting wire form an isosceles right triangle as shown below, antenna supporting wire Blitzer, Intermediate Algebra, 5e – Slide #33 Section 8.1
Isosceles Right Triangles in Application
CONTINUED this implies that the diagram can be represented as follows, using the “lengths within isosceles right triangles” principle.
70 ft.
antenna 70 ft.
supporting wire about 99 ft.
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 8.1
Pythagorean Theorem and the square root property in Application EXAMPLE:similar to #83 and #85 in homework Number 84 on page 573. A rectangular park is 4 miles long and 2 miles wide. How long is a pedestrian route that runs diagonally across the park?
SOLUTION Blitzer, Intermediate Algebra, 5e – Slide #35 Section 8.1
In Summary…
Solving Quadratic Equations by Completing the Square
1. If the coefficient of the second degree term is not 1, divide both sides by this coefficient 2. Isolate variable terms on one side 3. Complete the square by adding the square of half the coefficient of x to both sides 4. Factor the perfect square trinomial 5. Solve by applying the square root property Blitzer, Intermediate Algebra, 5e – Slide #36 Section 8.1
DONE
•
The relationships among the various sets of numbers.
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