Download M098 Carson Elementary and Intermediate Algebra 3e Section 11.1 Objectives

M098
Carson Elementary and Intermediate Algebra 3e
Section 11.1
Objectives
1.
2.
Use the square root principle to solve quadratic equations.
Solve quadratic equations by completing the square.
Vocabulary
Prior Knowledge
Solve quadratic equations by factoring.
New Concepts
1. Use the square root principle to solve quadratic equations.
Until now, all quadratic equations have been carefully chosen so that they could be solved by factoring.
Obviously, not every quadratic equation can be factored, so in Chapter 11, we learn some other methods
that will allow us to solve all quadratic equations.
2
Remember, we expect 2 solutions to quadratic equations because of the x term. Again, let’s look at some
examples.
2
If x = 25, then x = 5 or x = -5.
2
If y = 36, then y = 6 or y = -6.
How can we get the answers if the problem isn’t one that we can figure out in our head?
Notice the form: a perfect square isolated on the left side of the equation and no variable to the first
power. Then take the square root of both sides, remembering that there will be two answers. So we
always use ± in front of the radical on the right side.
Example 1: x 2  81
x 2   81
x=±9
The perfect square is isolated so take the square root of
both sides to remove the square. Remember the ± in front
of the radical.
There are two solutions: 9 and -9.
Example 2: a 2  20
a 2   20
a  2 5
The perfect square is isolated so take the square root of
both sides to remove the square. Remember the ± in front
of the radical.
There are two solutions: 9 and -9.
These are examples of the Square Root Property. Remember that solving an equation is a process of
undoing the operations on the variable. Addition and subtraction undo each other, multiplication and
division undo each other and now powers and roots undo each other.
V. Zabrocki 2011
page 1
M098
Carson Elementary and Intermediate Algebra 3e
Example 3: 
Section 11.1
2 2
8
m  
3
5
 2

 8
15  m 2   15  
3


 5


The perfect square is not isolate. Clear the fractions and
2
then isolate m .
5  2m 2  3 8 
 10m
2
 24
 24
 10
12

5
m2 
m2
m  
12

5
m  
The perfect square is isolated so take the square root of
both sides to remove the square. Remember the ± in front
of the radical.
12
5
m2  
5
Rationalize the radical and simplify.
5
60
25
m  
m
2 15
5
2 15
 2 15
or m 
5
5
There are two solutions.
The examples we’ve just looked at are the simplest kind: the perfect square was a monomial. The
process is the same if the base is a binomial.
Example 4:
a  62
a  62
 9
  9
The perfect square binomial is isolated, so take the square
root of both sides to remove the square. Remember the ±
in front of the radical.
a+6=±3
Solve for a.
a = -6 ± 3
Write the -6 in front of the ±. There are two solutions.
a = -6 + 3 or a = -6 – 3
a = -3 or a = -9
Notice the shorthand method (the ±) of writing the 2 cases. When you use this shortcut, always put the
number you bring to the right side in front of the ± sign.
V. Zabrocki 2011
page 2
M098
Carson Elementary and Intermediate Algebra 3e
5m  12
Example 5:
5m  12
 9  2
Isolate the perfect square binomial.
7
5m  12
Section 11.1
  7
The perfect square is isolated so take the square root of
both sides to remove the square. Remember the ± in front
of the radical.
5m  1   7
5m  1 
Solve for m. Place the -1 in front of the ±.
7
m 
1 7
5
m
1 7
5
There are two solutions.
or
m
1 7
5
t  52
Example 6:
t  52
 28
   28
The perfect square is isolated so take the square root of
both sides to remove the square. Remember the ± in front
of the radical.
t  5  2i 7
Simplify the radical. Rewrite the imaginary number using i.
t  5  2i 7
Solve for t.
t  5  2i 7
or
There are two solutions.
t  5  2i 7
2. Solve quadratic equations by completing the square.
As was mentioned earlier, very few quadratic equations can be factored. Even fewer of them come in the
form we saw in the last section: a perfect square equal to some number. However, ALL quadratic
equations can be changed into that form by a process called Completing the Square.
Let's start by looking at some perfect square trinomials:
2
2
(x + 6) = x + 12x + 36
2
2
(y - 4) = y - 8y + 16
2
2
(a + 3) = a + 6a + 9
V. Zabrocki 2011
page 3
M098
Carson Elementary and Intermediate Algebra 3e
Section 11.1
Here are some patterns you should notice:
1. the coefficient of the squared term is always 1.
2. The constant term is always equal to 1/2 the coefficient of the middle term squared.
2
1/2 (12) = 6
1/2 (8) = 4
1/2 (6) = 3
6 = 36
2
4 = 16
2
3 =9
3. Notice the constant term in the binomial on the left side. It is always equal to 1/2 the middle term
on the right side.
4. Notice the signs: The sign in the binomial is the same as the first sign of the trinomial. The last
sign of the trinomial is always positive.
You can use this knowledge to rewrite ANY quadratic equation.
Completing the Square Procedure
2
1.
The coefficient of squared term must be 1, so divide each term by the coefficient of the x term.
2.
Move the constant to the other side of the equation. (Leave space to add a term on each side.)
3.
Take half of the coefficient of the x term and square it.
4.
Add this amount to both sides of the equation.
5.
Rewrite the left side as a perfect square using one half of the coefficient of the x-term as the
number in the binomial and simplify the right side.
6.
Use the square root property. (Remember
7.
Finish the solution.
x2  10x  12  0
Example 7:
V. Zabrocki 2011
 !! )
x2  10x  12  0
The coefficient of squared term is already 1.
x 2  10x 
 12
Move the constant to the other side of the equation.
(Leave space to add a term on each side.)
1
10  5
2
52
Take half of the coefficient of the x term and square it.
 25
x 2  10x  25  12  25
Add this amount to both sides of the equation.
x  52
Rewrite the left side as a perfect square using one half of
the coefficient of the x-term as the number in the binomial
and simplify the right side.
 13
x  5   13
Use the square root property. (Remember
x  5  13
Finish the solution.
 !! )
page 4
M098
Carson Elementary and Intermediate Algebra 3e
2x 2  6x  5  0
Example 8:
The coefficient of squared term must be 1, so divide
each term by the coefficient of the x2 term.
2x 2
6x 5
0



2
2
2
2
x 2  3x 
5
 0
2
=
1
3  3
2
2
3
 
2
3
x 2  3x   
2
V. Zabrocki 2011
2

2
19
4

Move the constant to the other side of the equation.
(Leave space to add a term on each side.)
5

2
x 2  3x 
3

x  
2

Section 11.1
2

9
4
5 9

2 4
Take half of the coefficient of the x term and square it.
Add this amount to both sides of the equation.
Rewrite the left side as a perfect square using one half of
the coefficient of the x-term as the number in the binomial
and simplify the right side.
x
3
19
 
2
4
Use the square root property. (Remember
x
3
 19

2
2
Finish the solution.
x 
3

2
x 
3  19
2
 !! )
19
2
page 5