Transcript Quadratic Equations

10TH EDITION
COLLEGE ALGEBRA
LIAL
HORNSBY
SCHNEIDER
1.4 - 1
1.4
Quadratic Equations
Solving a Quadratic Equation
Completing the Square
The Quadratic Formula
Solving for a Specified Variable
The Discriminant
1.4 - 2
1.1
Quadratic Equation in One
Variable
An equation that can be written in the
2
form
ax  bx  c  0
where a, b, and c are real numbers
with a ≠ 0, is a quadratic equation.
The given form is called standard
form.
1.4 - 3
Second-degree Equation
A quadratic equation is a second-degree
equation.
This is an equation with a squared variable
term and no terms of greater degree.
x 2  25, 4x 2  4x  5  0, 3x 2  4x  8
1.4 - 4
Zero-Factor Property
If a and b are complex numbers with
ab = 0, then a = 0 or b = 0 or both.
1.4 - 5
USING THE ZERO-FACTOR
PROPERTY
Example 1
Solve 6x 2  7x  3
Solution:
6x 2  7x  3
6X  7X  3  0
(3x  1)(2x  3)  0
2
3x  1  0 or
2x  3  0
Standard form
Factor.
Zero-factor
property.
1.4 - 6
Example 1
USING THE ZERO-FACTOR
PROPERTY
Solve 6x 2  7x  3
Solution:
3x  1  0 or
3x  1
2x  3  0
or
2x  3
1
or
x
3
3
x
2
Zero-factor
property.
Solve each
equation.
1.4 - 7
Square-Root Property
A quadratic equation of the form x2 = k can
also be solved by factoring
x k
2
x k 0
2
 x  k  x  k   0
x  k  0 or x  k  0
x k
or x   k
Subtract k.
Factor.
Zero-factor property.
Solve each equation.
1.4 - 8
Square Root Property
If x2 = k, then
x k
or
x k
1.4 - 9
Square-Root Property
That is, the solution of
Both solutions
are real if k > 0,
and both are
imaginary if k < 0
If k < 0, we write
the solution set
as
i k 

x2  k
is
k,  k
or
 k

If k = 0, then this
is sometimes
called a double
solution.
1.4 - 10
Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
2
a. x  17
Solution:
By the square root property, the solution set
is
 17
1.4 - 11
USING THE SQUARE ROOT
PROPERTY
Example 2
Solve each quadratic equation.
b. x 2  25
Solution:
Since
1  i,
the solution set of x2 = − 25
is
5i .
1.4 - 12
Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
2
c. ( x  4)  12
Solution:
Use a generalization of the square root
property.
2
( x  4)  12
x  4   12
x  4  12
x  42 3
Generalized square
root property.
Add 4.
12  4 3  2 3
1.4 - 13
Solving A Quadratic Equation
By Completing The Square
To solve ax2 + bx + c = 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 equality symbol.
Step 3 Square half the coefficient of x, and add this square
to both sides of the equation.
Step 4 Factor the resulting trinomial as a perfect square
and combine like terms on the other side.
Step 5 Use the square root property to complete the
solution.
1.4 - 14
Example 3
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution
Step 1 This step is not necessary since a = 1.
Step 2
x2  4x  14
Add 14 to both
sides.
Step 3
x  4x  4  14  4
 1 ( 4)  4;
2

Step 4
2
( x  2)  18
2
2
add 4 to both sides.
Factor; combine
terms.
1.4 - 15
Example 3
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution
Step 4
( x  2)  18
Factor; combine terms.
Step 5
x  2   18
Square root property.
Take both
roots.
2
x  2  18
Add 2.
x  23 2
Simplify the radical.


The solution set is 2  3 2 .
1.4 - 16
USING THE METHOD OF
COMPLETING THE SQUARE a ≠ 1
Example 4
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
9x  12x  9  0
2
4
x  x  1 0
3
4
2
x  x  1
3
4
4
4
2
x  x   1
3
9
9
2
Divide by 9. (Step 1)
Add – 1. (Step 2)
2
4
 1  4  4


;
add
2  3  9
9
1.4 - 17
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
4
4
4
x  x   1
3
9
9
2
2
4
 1  4  4


;
add
2  3  9
9
2
 x  2  5


3
9

2
5
x  
3
9
Factor, combine
terms. (Step 4)
Square root property
1.4 - 18
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
2
5 Square root property
Solution
x  
3
9
2
5
x  i
3
3
2
5
x  i
3 3
a  i a
Quotient rule for
radicals
Add ⅔.
1.4 - 19
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
2
5
x  i
3 3
Add ⅔.
2
5 
i .
The solution set is  
3 3 
1.4 - 20
The Quadratic Formula
The method of completing the square can
be used to solve any quadratic equation. If
we start with the general quadratic equation,
ax2 + bx + c = 0, a ≠ 0, and complete the
square to solve this equation for x in terms
of the constants a, b, and c, the result is a
general formula for solving any quadratic
equation. We assume that a > 0.
1.4 - 21
Quadratic Formula
The solutions of the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are
b  b  4ac
x
.
2a
2
1.4 - 22
Caution Notice that the fraction bar
in the quadratic formula extends under
the –b term in the numerator.
b  b  4ac
x
.
2a
2
1.4 - 23
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Example 5
Solve x2 – 4x = –2
Solution:
x  4x  2  0
2
Write in standard
form.
Here a = 1, b = –4, c = 2
b  b2  4ac
x
2a
Quadratic formula.
1.4 - 24
Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = –2
Solution:
b  b  4ac
x
2a
2
The fraction
bar extends
under –b.
Quadratic formula.
( 4)  ( 4)2  4(1)(2)

2(1)
1.4 - 25
Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = –2
Solution:
( 4)  ( 4)  4(1)(2)

2(1)
2
The fraction
bar extends
under –b.
4  16  8

2
42 2

2
16  8  8  4 2  2 2
1.4 - 26
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Example 5
Solve x2 – 4x = –2
Solution:
42 2

2
2 2 2

2

Factor first,
then divide.
 2 2
16  8  8  4 2  2 2

Factor out 2 in the numerator.
Lowest terms.


The solution set is 2  2 .
1.4 - 27
Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution:
2x 2  x  4  0
Write in standard form.
(1)  (1)  4(2)(4)
x
2(2)
2
1 1 32

4
Quadratic formula;
a = 2, b = – 1, c = 4
Use parentheses and
substitute carefully to
avoid errors.
1.4 - 28
Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution:
1 1 32

4
1 31
x
1  i
4
1
31 
i .
The solution set is  
4 4 
1.4 - 29
Cubic Equation
The equation x3 + 8 = 0 that follows is called
a cubic equation because of the degree 3
term. Some higher-degree equations can
be solved using factoring and the quadratic
formula.
1.4 - 30
Example 7
SOLVING A CUBIC EQUATION
Solve x 3  8  0.
Solution
x 8 0
3
 x  2  x  2x  4  0
2
x  2  0 or x 2  2x  4  0
Factor as a sum of
cubes.
Zero-factor property
(2)  (2)  4(1)(4)
x  2 or x 
2(1)
2
Quadratic formula; a = 1, b = –2, c = 4
1.4 - 31
SOLVING A CUBIC EQUATION
Example 7
Solve x 3  8  0.
Solution
2  12
x
2
Simplify.
2  2i 3
x
2
Simplify the radical.
x

2 1 i 3
2

Factor out 2 in the
numerator.
1.4 - 32
Example 7
SOLVING A CUBIC EQUATION
Solve x 3  8  0.
Solution
x  1 i 3
Lowest terms


The solution set is 2,1 i 3 .
1.4 - 33
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve for the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
d 2
A
4
4A  d
4A
2
d

2
Goal: Isolate d,
the specified
variable.
Multiply by 4.
Divide by .
1.4 - 34
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution 4A
2
d

See the Note
following this
example.
4A
d 

Divide by .
Square root
property
1.4 - 35
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
4A
d 

 4A
d

Square root
property


Rationalize the
denominator.
1.4 - 36
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
 4A
d

 4A
d



Rationalize the
denominator.
Multiply numerators;
multiply
denominators.
1.4 - 37
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
 4A
d

Multiply numerators;
multiply
denominators.
2 A
d

Simplify.
1.4 - 38
Solving for a Specified Variable
Note In Example 8, we took both
positive and negative square roots.
However, if the variable represents a
distance or length in an application, we
would consider only the positive square
root.
1.4 - 39
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
Solution
(r  0), for t
r t  st  k  0 Write in standard form.
2
Now use the quadratic formula to find t.
b  b2  4ac
t
2a
1.4 - 40
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
(r  0), for t
Solution t  b  b  4ac
2
2a
(s)  (s)  4(r )(k )
t
2r
2
a = r, b = – s,
and c = – k
1.4 - 41
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
(r  0), for t
Solution
(s)  (s)  4(r )(k )
t
2r
a = r, b = – s,
and
c = –k
s  s  4rk
t
2r
Simplify.
2
2
1.4 - 42
The Discriminant
The Discriminant The quantity under the
radical in the quadratic formula, b2 – 4ac, is
called the discriminant.
b  b  4ac
x
2a
2
Discriminant
1.4 - 43
The Discriminant
Discriminant
Positive, perfect
square
Positive, but not
a perfect
square
Zero
Negative
Number of
Solutions
Type of
Solution
Two
Rational
Two
Irrational
One
(a double solution)
Rational
Two
Nonreal
complex
1.4 - 44
Caution The restriction on a, b, and c
is important. For example, for the equation
x  5x  1  0
2
the discriminant is b2 – 4ac = 5 + 4 = 9,
which would indicate two rational solutions
if the coefficients were integers. By the
quadratic formula, however, the two
solutions are irrational numbers,
53
x
2
1.4 - 45
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
a. 5x 2  2x  4  0
Solution
For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = –4.
The discriminant is
b2 – 4ac = 22 – 4(5)(–4) = 84
The discriminant is positive and not a perfect
square, so there are two distinct irrational
solutions.
1.4 - 46
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
b. x 2  10x  25
Solution
First write the equation in standard form as
x2 – 10x + 25 = 0. Thus, a = 1, b = –10, and
c = 25, and b2 – 4ac =(–10 )2 – 4(1)(25) = 0
There is one distinct rational solution, a “double
solution.”
1.4 - 47
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
c. 2x 2  x  1  0
Solution
For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so
b2 – 4ac = (–1)2 – 4(2)(1) = –7.
There are two distinct nonreal complex
solutions. (They are complex conjugates.)
1.4 - 48