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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
Quadratic Functions and
Equations
Quadratic Equations
The Quadratic Formula
Applications Involving Quadratic Equations
Studying Solutions of Quadratic Equations
Equations Reducible to Quadratic
Quadratic Functions and Their Graphs
More About Graphing Quadratic Functions
Problem Solving and Quadratic Functions
Polynomial and Rational Inequalities
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
8.1
Quadratic Equations

The Principle of Square Roots

Completing the Square

Problem Solving
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Principle of Square Roots
Let’s consider x2 = 25. We know that the
number 25 has two real-number square
roots, 5 and -5, the solutions of the
equation. Thus we see that square roots can
provide quick solutions for equations of the
type x2 = k.
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The Principle of Square Roots
For any real number k,
If x2 = k, then x  k or x   k .
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Slide 8- 5
Example
Solve 5x 2 = 15. Give exact solutions and
approximations to three decimal places.
Solution
5 x 2  15
x2  3
x  3 or x   3.
Isolating x2
Using the principle
of square roots
We often use the symbol  3 to represent both
solutions.
The solutions are  3, which round to 1.732 and
–1.732. The check is left to the student.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 6
Example
Solve 16x2 + 9 = 0.
Solution
2
16 x  9  0
2
x  9/16
x  9/ 4 or x   9/ 4
3
3
Recall that 1  i.
x  i or x   i.
2
2
3
The solutions are  i. The check is left to the
2
student.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 7
The Principle of Square Roots
(Generalized Form)
For any real number k and any algebraic
expression X,
If X 2 = k, then X  k or X   k .
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Example
Solve (x + 3)2 = 7.
Solution
( x  3)2  7
x  3  7 or x  3   7
x  3  7 or x  3  7.
The solutions are 3  7.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 9
Completing the Square
Not all quadratic equations can be solved as
in the previous examples. By using a
method called completing the square, we
can use the principle of square roots to solve
any quadratic equation.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 10
Example
Solve x2 + 10x + 4 = 0
Solution
x2  10 x  4  0
x2  10 x
x2
 4
+ 10x + 25 = –4 + 25
2
( x  5)  21
x  5  21 or x  5   21
Adding 25 to
both sides.
Factoring
Using the principle
of square roots
The solutions are 5  21.
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Slide 8- 11
To Solve a Quadratic Equation in x by
Completing the Square
1.Isolate the terms with variables on one side
of the equation, and arrange them in
descending order.
2.Divide both sides by the coefficient of x2 if
that coefficient is not 1.
3.Complete the square by taking half of the
coefficient of x and adding its square to
both sides.
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Slide 8- 12
4. Express the trinomial as the square of a
binomial (factor the trinomial) and simplify
the other side.
5. Use the principle of square roots (find the
square roots of both sides).
6. Solve for x by adding or subtracting on
both sides.
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Problem Solving
After one year, an amount of money P,
invested at 4% per year, is worth 104% of P,
or P(1.04). If that amount continues to earn
4% interest per year, after the second year the
investment will be worth 104% of P(1.04), or
P(1.04)2. This is called compounding
interest since after the first period, interest is
earned on both the initial investment and the
interest from the first period. Generalizing,
we have the following.
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The Compound Interest Formula
If the amount of money P is invested at
interest rate r, compounded annually, then in
t years, it will grow to the amount A given
by
A  P(1  r )t . (r is written in decimal notation.)
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Example
Jackson invested $5800 at an interest rate of r,
compounded annually. In 2 yr, it grew to $6765.
What was the interest rate?
Solution
1. Familiarize. We are already familiar with the
compound-interest formula.
2. Translate. The translation consists of
substituting into the formula:
t
A  P(1  r )
6765 = 5800(1 + r)2
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3. Carry out. Solve for r:
6765/5800 = (1 + r)2
 6765/5800  1  r
1  6765/5800  r
r  .08 or r  2.08
4. Check. Since the interest rate cannot negative, we
need only to check .08 or 8%. If $5800 were
invested at 8% compounded annually, then in 2 yr it
would grow to 5800(1.08)2, or $6765. The number
8% checks.
5. State. The interest rate was 8%.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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