chapter 8 quadratic equations and functions. § 8.1 the square root property and completing the...
TRANSCRIPT
Chapter 8Quadratic Equations and
Functions
§ 8.1
The Square Root Property and Completing the Square
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.1
Introduction
What we already know about quadratic equations…
A quadratic equation can be written in the standard form:
0,02 acbxax
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 #4 Section 8.1
The Square Root Property
The Square Root PropertyIf u is an algebraic expression and d is a nonzero real number, then has exactly two solutions:
Equivalently,
du 2
.or then , If 2 dududu
. then , If 2 dudu
This property says that when we take the square root of both sides of an equation, we get two roots. Don’t forget the
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.1
Using the Square Root Property
EXAMPLEEXAMPLE
Solve: .494 2 x
SOLUTIONSOLUTIONTo apply the square root property, we need a squared expression by itself on one side of the equation. We can get by itself if we divide both sides by 4.
This is the given equation.
2x
494 2 x
Divide both sides by 4.4
492 x
Apply the square root property.4
49x
Simplify.5.32
7
4
49x
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.1
Using the Square Root Property
494 2 x
The solutions are 3.5 and -3.5. The solution set is {3.5,-3.5}.
CONTINUECONTINUEDD
Check 3.5:
494 2 x
Check -3.5:
495.34 2 495.34 2
4925.124 4925.124 4949 4949
?
? ?
?
true true
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.1
Using the Square Root Property
EXAMPLEEXAMPLE
Solve: .0494 2 x
SOLUTIONSOLUTION
To solve by the square root property , we isolate the squared expression on one side of the equation.
This is the given equation.494 2 x
Divide both sides by 4.4
492 x
Apply the square root property.4
49x
0494 2 x
Subtract 49 from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.1
Using the Square Root Property
0494 2 x
The solutions are 3.5i and -3.5i. The solution set is {3.5i,-3.5i}.
CONTINUECONTINUEDD
Check 3.5i:0494 2 x
Check -3.5i:
?
? ?
?
iiix 5.32
7
4
49
14
49x
i 1
0495.34 2 i 0495.34 2 i
04925.124 2 i 04925.124 2 i
049125.124 049125.124 00 00
? ?
truetrue
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.1
Using the Square Root Property
EXAMPLEEXAMPLE
Solve by the square root property: .3623 2 xSOLUTIONSOLUTION
This is the given equation.
Apply the square root property.
Subtract 2 from both sides of each equation.
Divide both sides by 3.
3623 2 x
122 2 x
122or 122 xx
122or 122 xx
Rewrite radicands.342or 342 xx
Simplify.322or 322 xx
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.1
Using the Square Root Property
Check : Check :322 322
3623 2 x 3623 2 x
36232232 3623223
2
CONTINUECONTINUEDD
363232 36323
2
3632322 36323
22
36343 36343 3636 3636
The solutions are . The solution set is322 . 322
?
?
?
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truetrue
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.1
Completing the Square
Completing the Square
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,
bxx 2
2
2
b
.22
222
bx
bbxx
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.1
Completing the Square
EXAMPLEEXAMPLE
What term should be added to the binomial so that it becomes a perfect square trinomial? Write and factor the trinomial.
xx 102 SOLUTIONSOLUTION
xx 102
Add Add 25 to complete the square.
.52
10 22
22 52510 xxx
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.1
Completing the Square
EXAMPLEEXAMPLE
Solve by completing the square: .0352 2 xx
SOLUTIONSOLUTION
This is the given equation.
0352 2 xx
Divide both sides by 2.02
3
2
52 xx
Add 3/2 to both sides.2
3
2
52 xx
Complete the square: Half of 5/2 is16
25
2
3
16
25
2
52 xx
.16
25
4
5 and ,
4
52
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.1
Completing the Square
Factor and simplify.25
49
16
25
16
24
4
52
x
CONTINUECONTINUEDD
Apply the square root property.25
49
4
5x
5
7
4
5x
5
7
25
49
25
49
Split into two equations and subtract 5/4 from both sides of both equations.
4
5
5
7or
4
5
5
7 xx
Get common denominators.20
25
20
28or
20
25
20
28 xx
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.1
Completing the Square
CONTINUECONTINUEDD
Simplify.20
53or
20
3 xx
The solutions are , and the solution set is20
53 and
20
3 .
20
53,
20
3
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.1
Completing the Square in Application
A Formula for Compound InterestSuppose 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
.1 trPA
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.1
Completing the Square in Application
EXAMPLEEXAMPLE
Use the compound interest formula, , to find the annual interest rate r.
In 2 years, an investment of $80,000 grows to $101,250.
SOLUTIONSOLUTION
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.
trPA 1
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 8.1
Completing the Square in Application
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.
trPA 1
CONTINUECONTINUEDD
Use the compound interest formula.
21000,80250,101 r Substitute the given values.
21000,80
250,101r Divide both sides by 80,000.
2164
81r Simplify the
fraction.
64
811or
64
811 rr Apply the square root
property.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.1
Completing the Square in Application
CONTINUECONTINUEDD
8
91or
8
91 rr
8
9
64
81
64
81
18
9or 1
8
9 rr Subtract 1 from both sides.
8
17or
8
1 rr Simplify.
Because the interest rate cannot be negative, we reject -17/8. Thus, the annual interest rate is 1/8 = 0.125 = 12.5%.
We can check this answer using the formula . If $80,000 is invested for 2 years at 12.5% interest, compounded annually, the balance in the account is
trPA 1
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.1
Completing the Square in Application
CONTINUECONTINUEDD
Because this is precisely the amount given by the problem’s conditions, the annual interest rate is, indeed, 12.5% compounded annually.
.250,101$125.1000,80$125.01000,80$ 22 A
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.1
Isosceles Right Triangles
Lengths Within Isosceles Right TrianglesThe 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 #22 Section 8.1
Isosceles Right Triangles in Application
EXAMPLEEXAMPLE
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?
SOLUTIONSOLUTION
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,
supporting wireantenna
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.1
Isosceles Right Triangles in Application
this implies that the diagram can be represented as follows, using the “lengths within isosceles right triangles” principle.
supporting wireantenna
CONTINUECONTINUEDD
70 ft.
70 ft.
70 ft.2
Therefore the supporting wire must be 70 feet which equals about 99 ft.
2
Blitzer, Intermediate Algebra, 5e – Slide #24 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 coefficient2.Isolate variable terms on one side3.Complete the square by adding the square of half the coefficient of x to both sides4.Factor the perfect square trinomial5.Solve by applying the square root property