Download 4.5 Completing the Square

Algebra 2
Lesson 4-5
Example 1 Equation with Rational Roots
Solve 2x2 – 36x + 162 = 32 by using the Square Root Property.
2x2 – 36x + 162 = 32
2(x2 – 18x + 81) = 2(16)
x2 – 18x + 81 = 16
(x – 9)2 = 16
x – 9 = ± 16
Original equation
Factor out the GCF.
Divide each side by 2.
Factor the perfect trinomial square.
Square Root Property
x – 9 = ±4
x=9±4
x=9+4
or x = 9 – 4
x = 13
x=5
16 = 4
Add 9 to each side.
Write as two equations.
Solve each equation.
The solution set is {5, 13}. You can check this result by using factoring to solve the
original equation.
Example 2 Equation with Irrational Roots
Solve x2 + 10x + 25 = 108 by using the Square Root Property.
x2 + 10x + 25 = 108
(x + 5)2 = 108
x + 5 = ± 108
Original equation
Factor the perfect square trinomial.
Square Root Property
x = –5 ±6 3
x = –5 + 6 3 or x = –5 – 6 3
x ≈ 5.4
x ≈ –15.4
Add –5 to each side; 108 = 6 3
Write as two equations.
Use a calculator.
The exact solutions of this equation are –5 – 6 3 and –5 + 6 3 . The approximate
solutions are –15.4 and 5.4. Check these results by finding and graphing the related
quadratic function.
x2 + 10x + 25 = 108
x2 + 10x – 83 = 0
y = x2 + 10x – 83
Original equation
Subtract 108 from each side.
Related quadratic function.
CHECK Use the ZERO function of a graphing
calculator. The approximate zeros of the related
function are –15.4 and 5.4.
Example 3
Complete the Square
5
Find the value of c that makes x2 – x + c a perfect square. Then write the trinomial
3
as a perfect square.
5
5
3
3
Find one half of .
Step 1
•
1
2
=
5
6
2
 5
25
 6  = 36
Step 2
Square the result of Step 1
Step 3
Add the result of Step 2 to x2 – x
5
25
3
36
x2 – x +
5
3
2

5
The trinomial x – x +
can be written as  x   .
3
36
6

5
2
25
Example 4 Solve an Equation by Completing the Square
Solve x2 – 7x – 44 = 0 by completing the square.
x2 – 7x – 44 = 0
x2 – 7x = 44
x – 7x +
2
49
4
Notice that x2 – 7x – 44 is not a perfect square.
Rewrite so the left side is of the form x2 + bx.
= 44 +
2
 7
49
49
Since   = , add
to each side.
 2
4
4
49
4
2

7  225
 x  2  = 4
7
225
2
4
x – =
Square Root Property
x – =
7
15
225
2
2
4
7

x=
x=
Write the left side as a perfect square by factoring.
7
2
x = 11
+
15
2
2
15
Add
2
or x =
7
2
x = –4
–
15
2
=
7
2
15
2
to each side.
Write as two equations.
The solution set is {-4, 11}.
You can check this result by using factoring to solve the original equation.
Example 5 Equation with a ≠ 1
Solve 3x2 + 4x – 7 = 0 by completing the square.
Notice that 3x2 + 4x – 7 is not a perfect square.
3x2 + 4x – 7 = 0
Divide by the coefficient of the quadratic term, 3.
4
7
=0
x2 + x –
3
x2 +
2
x +
4
3
x+
3
4
x =
3
4
9

2
 x  3 
x+
=
2
=
2
3
7
Add
3
7
+
3
2
5
3
3
x=– +
x=1
2
9
9
5
Square Root Property
3
2
5
3
3
x =– ±
to each side.
3
  4

4
4
Since     2 = , add to each side.
9
9
  3

Write the left side as a perfect square by factoring.
Simplify the right side.
4
25
=±
7
2
Add – to each side.
3
2
5
3
3
or x = – –
x=–
7
3
Write as two equations.
 7 
The solution set is  , 1 .
 3 
Example 6
Equation with Imaginary Solutions
2
Solve 4x – 2x + 7 = 0 by completing the square.
4x2 – 2x + 7 = 0
1
7
2
4
x2 – x +
=0
7
2
4
1
1
2
16
x – x+
Divide by the coefficient of the quadratic term, 4.
1
x2 – x = –
2
Notice that 4x2 – 2x + 7 is not a perfect square.
Rewrite so the left side is of the form x2 + bx.
7
1
4
16
=– +
2
 1

1
1
Since    2 = , add
to each side.
 2

16
16
2

1
27
=–
x



16
4
Write the left side as a perfect square by factoring.
Simplify the right side.
1
x– = 
4
27
16
1
3i 3
4
4
x– =±
x=
1
4
±
Square Root Property
1 = i
3i 3
4
Add
1
4
to each side.
 1 3i 3 1 3i 3 
, 
The solution set is  
 . Notice that these are imaginary solutions.
4 4
4 
 4