Download Solving Quadratic Equations by Graphs and

Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
x 2 ­ 8 x + 15 = 0
Consider the equation
• Show (numerically) that x =
5 is a solution.
.
There is also another solution to
the equation.
Use the table to help find this
second answer.
x
y
x 2 ­ 8 x + 15
1
(1)2 – 8(1) + 15
8
2
(2)2 – 8(2) + 15
3
3
(3)2 – 8(3) + 15
0
4
(4)2 – 8(4) + 15
-1
5
(5)2 – 8(5) + 15
0
6
(6)2 – 8(6) + 15
3
Sample:
(5)2 – 8(5) + 15
25 – 40 + 15
-15 + 15
0
Using the table above, plot the points (x, y) to sketch the graph of
y = x 2 ­ 8 x + 15
.
• Which two points represent the solutions to the original equation?
Write down their coordinates.
(3,0) and (5,0)
•
Where are both of these points located?
Both points are on the x-axis.
Here, we found both solutions to a quadratic equation using a method like trial-and-error.
However, there are other (more efficient) methods for solving such quadratic equations:
Part I: Graphs and Tables
Part II: Factoring
Part III: Quadratic Formula
In all these cases, we use the following terms to describe the solutions to quadratic equations.
x 2 ­ 6x + 8 = 0
©2012, TESCCC
09/06/12
page 1 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
When we are dealing with general equations,
then the answers are called
When we are dealing with functions,
y = x 2 ­ 6x + 8
such as
,
then the answers are called
solutions or roots
zeros or x-intercepts
Here, we could say,
x 2 ­ 6x + 8 = 0
“The solutions to
are x = 2 and x = 4.”
Here, we could say,
y = x 2 ­ 6x + 8
“The zeros of the function
occur at x = 2 and x = 4.”
like this one, with x’s (but no y’s),
©2012, TESCCC
09/06/12
page 2 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
Part I: Graphs and Tables
Use a graphing calculator to complete the chart.
WINDOW
:
Factored
Equation
Ex.
Expanded
Equation
y = (x – 2)(x +
2)
2
y = ( x ­ 2)( x + 2) y = x + 2x –
2x – 4
Graph
(Sketch)
Xmin= -9.4
Xmax= 9.4
Zeros of
Function
x = -2
x=2
Ymin= -12
Ymax= 12
Check
(with Table)
x
-2
0
2
y
0
-4
0
x
-1
4
y
0
0
x
-3
-6
y
0
0
x
5
y
0
x
-2
2.5
y
0
0
x
-4
1.5
y
0
0
y = x2 – 4
1)
2)
y = ( x ­ 4)( x + 1)
y = x2 – 3x – 4
y = ( x + 3)( x + 6) y = x2 + 9x +
18
x = -1
x=4
x = -3
x = -6
y = ( x ­ 5)( x ­ 5)
3)
,
or
y = ( x ­ 5) 2
y = x2 – 10x +
25
4)
y = ­ 2( x ­ 2.5)( x +y2=) -2x2 + x +
10
5)
y = 2(x – 1.5)
(x + 4),
or
y = (2x – 3)(x +
4)
©2012, TESCCC
y = 2 x 2 + 5 x ­ 12
09/06/12
x=5
x = -2
x = 2.5
x = -4
x = 1.5
page 3 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
6)
©2012, TESCCC
None
(no real
solutions)
y = x2 + 4
09/06/12
x
-1
0
1
y
5
4
5
page 4 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
Questions:
7)
How is each factored equation related to the zeros of its related function?
Answers may vary.
Sample: The factors have numbers that are the opposites of the zeros.
Sample: If n is a zero of the function, then (x – n) is a factor.
y = ( x ­ 5)( x ­ 5)
8)
Equation #3 used the same factor twice:
. How does this affect the graph of its
related function?
Answers may vary.
Sample: The graph only has one x-intercept.
Sample: The graph touches (but does not cross) the x-axis at the point.
y = ­ 2( x ­ 2.5)( x + 2)
9)
Equation #4 used a negative coefficient in front of its factors:
this affect the graph of its related function?
Answers may vary.
Sample: The negative number made the graph point down instead of up.
Sample: The negative coefficient reflected the graph over the x-axis.
. How does
Think About It:
The Fundamental Theorem of Algebra states that a polynomial equation of degree n has precisely n
complex roots.
• What is the degree of a quadratic equation? 2
•
How many complex roots should it have?
It should have two complex roots.
Study the graphs of the quadratic functions below. Explain how their related equations can each have
“two complex roots.”
Explain:
Two real roots.
The two solutions are where it
crosses the x-axis and can be
rational or irrational.
©2012, TESCCC
Explain:
One real “double root.”
The two solutions are equal to
the same number. This is why it
is called a double root.
09/06/12
Explain:
No real solutions.
Solutions cannot be real
numbers because it does not
cross the axis. These solutions
involve imaginary numbers, but
are still considered “complex.”
page 5 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
©2012, TESCCC
09/06/12
page 6 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
Sample Problems
Complete the tables to determine the zeros of each quadratic function.
A)
B)
y = x 2 ­ 2x ­ 3
x
-2
-1
0
1
2
3
4
y
5
0
-3
-4
-3
0
5
Zeros:
x = -1, x = 3
f ( x ) = 2x 2 ­ 9 x + 7
x
1
1.5
…
3
3.5
4
4.5
y
0
-2
…
-2
0
3
7
Zeros:
x = 1, x = 3.5
Use a calculator to sketch the graph of each function. Then tell where the x-intercepts occur.
y = x 2 ­ 4x ­ 5
y = 2 x 2 + 24 x + 72
y = x 2 ­ 3x + 5
C)
Graph:
D)
Graph:
E)
Graph:
x-intercepts:
x = -1, x = 5
x-intercepts:
x = -6 (double root)
x-intercepts:
None (no real solutions)
F)
Write down the zeros of the
quadratic function graphed
here.
Work backward and use the
zeros to write the two factors
of the function.
Multiply the factors to find an
“expanded” equation for the
function.
©2012, TESCCC
Zeros:
x = -3 and x = 2
Factors:
(x + 3) and (x – 2)
Function:
y = x2 + x – 6
09/06/12
page 7 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
©2012, TESCCC
09/06/12
page 8 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
Part II: Factoring
Follow these steps to solve quadratic equations by factoring.
1. Set each equation equal to zero.
2. Factor.
3. Set each factor equal to zero and solve.
4. Check solutions by graphing the representative function and comparing zeros of the
function.
NOTE: Factoring cannot be used to find irrational or imaginary solutions.
Sample Problems
A)
x2 ­ x = 6
x2 – x – 6 = 0
(x – 3)(x + 2) = 0
{-2, 3}
B)
2x 2 = 5 ­ 9 x
2x2 + 9x – 5 = 0
(x + 5)(2x – 1) = 0
{-5, ½ }
C)
2 x 2 ­ 4 x = ­ 8 x + 16
2x2 + 4x – 16 = 0
2(x + 4)(x – 2) = 0
{-4, 2}
©2012, TESCCC
09/06/12
page 9 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
Solving Quadratic Equations by Graphs and Factoring KEY
PRACTICE PROBLEMS
Find the zeros of each function using tables and graphs.
y=
1)
Table:
1
2
x
1
2
3
4
5
6
y = x 2 ­ 14 x + 49
2)
Graph (sketch):
3)
Graph (sketch):
Zeros:
x = 2.75, x = 6
Zeros:
x = 7 (double root)
y
7.5
4
1.5
0
-0.5
0
Zeros:
x = 4, x = 6
4)
y = 4 x 2 ­ 35 x + 66
x 2 ­ 5 x + 12
Write down the zeros of the
quadratic function graphed
here.
Work backward and use the
zeros to write the two factors of
the function.
Multiply the factors to find an
“expanded” equation for the
function.
Zeros:
x = -4 and x = 1
Factors:
(x + 4) and (x – 1)
Function:
y = x2 + 3x – 4
Find the solutions to the quadratic equations by factoring. Check all possible solutions.
5)
x 2 + 6x + 9 = 0
6)
{-3} double root
7)
2x 2 + 7x + 3 = 0
−
{-3,
1
2
©2012, TESCCC
x 2 + 3 x = 10
{-5, 2}
8)
}
3x 2 + x = 2
{-1,
09/06/12
2
3
}
page 10 of 11
Algebra 2
HS Mathematics
Unit: 05 Lesson: 02
©2012, TESCCC
09/06/12
page 11 of 11