Transcript Completing the Square

13.3 Completing the Square
• Objective: To complete a square for a quadratic equation
and solve by completing the square
Steps to complete the square
• 1.) You will get an expression that looks like this:
AX²+ BX
• 2.) Our goal is to make a square such that we have
(a + b)² = a² +2ab + b²
• 3.) We take ½ of the X coefficient
(Divide the number in front of the X by 2)
• 4.) Then square that number
To Complete the Square
x2 + 6x
• Take half of the coefficient of ‘x’ 3
• Square it and add it 9
x2 + 6x + 9 = (x + 3)2
Complete the square, and show what the perfect
square is:
x  12x x  12x  36
2
2
y  14 y
y  14 y  49
y  10 y
y  10y  25
2
2
x  5x
2
2
2
25
x  5x 
4
2
x  6
2
 y  7
2
 y  5
2
5

x 
2

2
To solve by completing the square
• If a quadratic equation does not factor we can solve it by
two different methods
• 1.) Completing the Square (today’s lesson)
• 2.) Quadratic Formula (Next week’s lesson)
Steps to solve by completing the square
1.) If the quadratic does not factor, move the
constant to the other side of the equation
Ex: x²-4x -7 =0
x²-4x=7
2.) Work with the x²+ x side of the equation and
complete the square by taking ½ of the coefficient
of x and squaring
Ex. x² -4x
4/2= 2²=4
3.) Add the number you got to complete the square to
both sides of the equation
Ex: x² -4x +4 = 7 +4
4.)Simplify your trinomial square
Ex: (x-2)² =11
5.)Take the square root of both sides of the equation
Ex: x-2 =±√11
6.) Solve for x
Ex: x=2±√11
Solve by Completing the Square
x  6 x  16  0
2
x  6 x  16
2
+9
+9
x  6 x  9  25
2
 x  3  25
 x  3  5
x  3  5 x  8 x  2
2
Solve by Completing the Square
x  22 x  21  0
2
x  22 x  21
2
+121
+121
x  22 x  121  100
2
 x  11  100
 x 11  10
x  11  10 x  21 x  1
2
Solve by Completing the Square
x  2x  5  0
2
x  2x  5
2
+1
+1
x  2x 1  6
2
 x  1  6
 x 1   6
2
x  1 6
Solve by Completing the Square
x  10 x  4  0
2
x  10 x  4
2
+25
+25
x  10 x  25  29
2
 x  5  29
2
x  5   29
x  5  29
Solve by Completing the Square
x  8 x  11  0
2
x  8 x  11
2
+16
+16
x  8 x  16  5
2
 x  4  5
 x  4   5
2
x  4  5
Solve by Completing the Square
x  6x  4  0
2
x  6 x  4
2
+9
+9
x  6x  9  5
2
x  3  5
x  3   5
2
x  3 5
The coefficient of
2 x  3x  3  0
2
2
2
2
2
3
3
x  x 0
2
2
2
x
must be “1”
2
3  33

x  
4  16

2
3
33
x 
4
16
3
3 3 2 3
2
x  x
4
33
3
33
2
2 2
x   3 33
x

4
4
16
3
9
3
9
2
x 
2
x
16


2 16
4
The coefficient of
3x  12 x  1  0
1
2
x  4x  
3
2
x
must be “1”
2
1
x  4 x  4  4 
3
2
 x  2
2
11

3
11
x2  
3
11
x  2
3
33
x  2
3
3

3
6
x  32 
33
3
6  33
x
3