Title: Section 1'4 Quadratic Equations
1Section 1.4 Quadratic Equations
2Definition 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.
3Examples
- Quadratic Not Quadratic
- x2 36 x 25
- 5x2 5x ? 7 0 5x ? 8 0
- 3x2 4x ? 8 2x 4x ? 16
- x3 8 0
4Solving Quadratic Equations
- The following methods can be used to solve
- quadratic equations
- Zero-Factor Property
- Square Root Property
- Completing the Square
- Quadratic Formula
5The 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.
6Solving 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.
7Example
- 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
8Solution 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
9Example
- (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
10The 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 12Completing 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.
13Completing 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.
14Example 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
15Example a ? 1
Divide the equation by 6.
16Example a ? 1 continued
17Quadratic Formula
- The solutions of the quadratic equation ax2 bx
c 0, where a ? 0 are
18Example
(real solutions)
19Example
(nonreal complex solutions)
20Example
21Solving for a Variable that is Squared
- Solve the given formula for r, V ? r2h
Disregard the negative solution since r can not
be negative.
22The Discriminant
23Examples
- 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
24Section 1.5
Applications and Modeling with Quadratic Equations
25Problem 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.
26Example (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?
27Solution
- 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.
28Solution continued
- Step 3 Write an equation. V lwh
- Volume length ? width ? height
- 1435 (3x ? 10)(x ? 10)(5)
- Step 4 Solve the equation.
29Solution 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
30The 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
31Example
- 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.
32Example 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
33Example continued
34Example 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.
35Example (page 128)
36Example (cont.)
37Modeling 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.
38Modeling 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?
39Solution
- a)
- The predictions are greater than the actual
figures of 3.4 and 3.8 million.
For 2000, x 10
For 2001, x 11
40Solution 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.
41Homework
- 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