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Section 1'4 Quadratic Equations

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The bar graph shows the sales of SUV's in the United States, in millions. ... S = .016x2 .124x .787 models sales of SUV's from 1990 to 2001, where ... – PowerPoint PPT presentation

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Title: Section 1'4 Quadratic Equations


1
Section 1.4 Quadratic Equations
2
Definition of a Quadratic Equation
  • A quadratic equation in x is an equation that can
    be written in the standard form
  • ax2 bx c 0
  • where a, b, and c are real numbers with a not
    equal to 0. A quadratic equation in x is also
    called a second-degree polynomial equation in x.

3
Examples
  • Quadratic Not Quadratic
  • x2 36 x 25
  • 5x2 5x ? 7 0 5x ? 8 0
  • 3x2 4x ? 8 2x 4x ? 16
  • x3 8 0

4
Solving Quadratic Equations
  • The following methods can be used to solve
  • quadratic equations
  • Zero-Factor Property
  • Square Root Property
  • Completing the Square
  • Quadratic Formula

5
The Zero-Factor Property
  • 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.

6
Solving a Quadratic Equation using the
Zero-Factor Property
  • If necessary, rewrite the equation in the form
    ax2 bx c 0, moving all terms to one side,
    thereby obtaining zero on the other side.
  • Factor.
  • Apply the zero-product principle, setting each
    factor equal to zero.
  • Solve the equations in step 3.
  • Check the solutions in the original equation.

7
Example
  • Solve 2x2 7x 4 by factoring and then using
    the zero-factor property
  • Step 1 Move all terms to one side and obtain
    zero on the other side. Subtract 4 from both
    sides and write the equation in standard form.
  • 2x2 7x - 4 4 - 4
  • 2x2 7x - 4 0
  • Step 2 Factor.
  • 2x2 7x - 4 0
  • (2x - 1)(x 4) 0

8
Solution cont.
  • Steps 3 and 4 Set each factor equal to zero
    and solve each resulting equation.
  • 2 x - 1 0 or x 4 0
  • 2 x 1 x -4
  • x 1/2
  • Steps 5 check your solution

9
Example
  • (2x -3)(2x 1) 5
  • 4x2 - 4x - 3 5
  • 4x2 - 4x - 8 0
  • 4(x2-x-2)0
  • 4(x - 2)(x 1) 0
  • x - 2 0, and x 1 0
  • So x 2, or -1

10
The Square Root Property
  • If x is an algebraic expression and k is a
    positive real number, then x2 k has exactly two
    solutions.
  • If x2 k, then
  • Equivalently,
  • If x2 k, then

11
  • Examples

12
Completing the Square
  • To solve ax2 bx c 0, a ? 0, by completing
    the square
  • Step 1 If a ? 1, divide both sides of the
    equation by a.
  • Step 2 Rewrite the equation so that the
    constant term is alone on one side of the
    equal sign.
  • Step 3 Divide the b term by 2, then square it
    and add this to both sides of the equation.

13
Completing the Square continued
  • Step 4 Factor the resulting trinomial as a
    perfect square and combine terms on the
    other side.
  • Step 5 Use the square root property to
    complete the solution.

14
Example a 1
  • x2 ? 6x ? 12 0
  • Step 1 Not necessary since a 1.
  • Step 2 x2 ? 6x 12
  • Step 3 x2 ? 6x 9 12 9
  • Step 4 (x ? 3)2 21
  • Step 5

15
Example a ? 1
  • Complete the square.

Divide the equation by 6.
16
Example a ? 1 continued
  • The solution set is

17
Quadratic Formula
  • The solutions of the quadratic equation ax2 bx
    c 0, where a ? 0 are

18
Example
(real solutions)
19
Example
  • Solve 3x2 x ? 5.

(nonreal complex solutions)
20
Example
21
Solving for a Variable that is Squared
  • Solve the given formula for r, V ? r2h

Disregard the negative solution since r can not
be negative.
22
The Discriminant
23
Examples
  • Determine the number of distinct solutions and
    tell whether they are rational, irrational, or
    nonreal complex numbers.

3x2 ? x 2 0 b2 ? 4ac (?1)2?(4)(2)(3)
?23 Two nonreal complex solutions
x2 ? 12x ?36 x2 ? 12x 36 0 b2 ? 4ac
(?12)2?(4)(36) 0 One rational double solution
3x2 x ? 5 0 b2 ? 4ac 12?(4)(3)(?5) 61 Two
irrational solutions
24
Section 1.5
Applications and Modeling with Quadratic Equations
25
Problem Solving
  • When solving problems that lead to quadratic
    equations, we may get a solution that does not
    satisfy the physical constraints of the problem.
    For example, if x represents a width and two
    solutions of the quadratic equation are ?9 and 1,
    the value ?9 must be rejected since a width must
    be a positive number.

26
Example (page 126)
  • A piece of machinery is capable of producing
    rectangular sheets of metal such that the length
    is three times the width. Furthermore,
    equal-sized squares measuring 5 in. on a side can
    be cut from the corners so that the resulting
    piece of metal can be shaped into an open box by
    folding up the flaps. If specifications call for
    the volume of the box to be 1435 in.3, what
    should the dimensions of the original piece of
    metal be?

27
Solution
  • Step 1 Read the problem. We must find the
    dimensions of the original piece of metal.
  • Step 2 Assign a variable. The length is 3 times
    the width, let x width and 3x length
  • The box is formed by cutting 5 5 10 inches
    from both the length and width. The length and
    width of the box are shown on the figures.

28
Solution continued
  • Step 3 Write an equation. V lwh
  • Volume length ? width ? height
  • 1435 (3x ? 10)(x ? 10)(5)
  • Step 4 Solve the equation.

29
Solution continued
  • Step 5 State the answers. Only 17 satisfies the
    restriction. Thus, the dimensions of the
    original piece of metal should be 17in. by 3(17)
    51 in.
  • Step 6 Check. The length of the bottom of the box
    is 51 ? 2(5) 41 in. The width is
  • 17 ? 2(5) 7 in. The height is 5 in. So,
    the volume of the box is
  • 41(7)(5) 1435 in.3

30
The Pythagorean Theorem
  • In a right triangle, the sum of the squares of
    the lengths of the legs is equal to the square of
    the length of the hypotenuse.
  • a2 b2 c2

31
Example
  • A kite is flying on 70 feet of string. Its
    vertical distance from the ground is 15 feet more
    than its horizontal distance from the person
    flying it. Assuming that the string is being held
    at ground level, find its horizontal distance
    from the person and its vertical distance from
    the ground.

32
Example continued
  • Step 1 Read the problem. We are finding
    the horizontal and vertical distance.
  • Step 2 Assign a variable.
  • Let x the horizontal distance
  • Let x 15 the vertical distance
  • Step 3 Write an equation.
  • x2 (x 15)2 702

33
Example continued
  • Step 4 Solve

34
Example continued
  • Step 5 State the answer. Since x represents a
    length, the horizontal distance could only be
    41.43 feet. The vertical distance would be 41.43
    15 56.43 feet.
  • Step 6 Check. The lengths 41.43, 56.43 and 70
    satisfy the Pythagorean theorem.

35
Example (page 128)
36
Example (cont.)
37
Modeling with Quadratic Equations (page 129)
  • The bar graph shows the sales of SUVs in the
    United States, in millions. The quadratic
    equation S .016x2 .124x .787 models sales
    of SUVs from 1990 to 2001, where S represents
    sales in millions and x 0 represents 1990, x
    1 represents 1991, and so on.
  • a) Use this model to predict sales in 2000 and
    2001. Compare the results to the actual figures
    of 3.4 million and 3.8 million from the graph.

38
Modeling with Quadratic Equations continued
  • b) According to the model, in what year do sales
    reach 3 million? (Round down to the nearest
    year.) Is the result accurate?

39
Solution
  • a)
  • The predictions are greater than the actual
    figures of 3.4 and 3.8 million.

For 2000, x 10
For 2001, x 11
40
Solution continued
  • b)
  • Reject the negative solution and round 8.5
    down to 8 The year 1998 corresponds to x 8. The
    model is a bit misleading, since SUV sales did
    not reach 3 million until 2000.

41
Homework
  • 1.4 page 123
  • 13-27odd, 31-41odd, 45-53odd, 59-67odd,
    71-79odd
  • 1.5 page 130
  • 5, 7, 9, 17, 19, 21, 25, 27, 31
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