Opening Square Brackets
Some quadratic functions can be written as perfect squares.
(x + 1)2 = (x + 1)(x + 1)
= x2 + x + x + 1
= x2 + 2x + 1
(x + 3)2 = (x + 3)(x + 3)
= x2 + 3x + 3x + 9
= x2 + 6x + 9
Opening Square Brackets
Some quadratic functions can be written as perfect squares.
(x + 5)2 = (x + 5)(x + 5)
= x2 + 5x + 5x + 25
= x2 + 10x + 25
(x – 2)2 = (x – 2)(x – 2)
= x2 – 2x – 2x + 4
= x2 – 4x + 4
Opening Square Brackets
Some quadratic functions can be written as perfect squares.
(x + 1)2 = x2 + 2x + 1
(x + 3)2 = x2 + 6x + 9
(x + 5)2 = x2 + 10x + 25
(x – 2)2 = x2 – 4x + 4
Opening Square Brackets
Some quadratic functions can be written as perfect squares.
(x + 1)2 = x2 + 2x + 1
(x + 3)2 = x2 + 6x + 9
(x + 5)2 = x2 + 10x + 25
(x – 2)2 = x2 – 4x + 4
Then what would (x + 9)2 be?
x2 + 18x + 81
Then what would (x + 11)2 be? x2 + 22x + 121
Then what would (x – 7)2 be?
x2 – 14x + 49
Then what would (x + a)2 be?
x2 + 2ax + a2
Then what would (x – a)2 be?
x2 – 2ax + a2
Completing The Square
Using this idea we can factorise some quadratic functions into perfect squares.
x2 + 8x + 16
x2 + 10x + 25
(x + 4)2
(x + 5)2
x2 + 6x + 9
x2 + 14x + 49
(x + 3)2
(x + 7)2
Completing The Square
Using this idea we can factorise some quadratic functions into perfect squares.
x2 – 12x + 36
(x – 6)2
When we write
expressions in this
form it is known as
completing the
square.
x2 – 24x + 144
(x – 12)2
x2 – 20x + 100
(x – 10)2
x2 – 2x + 1
(x – 1)2
Completing The Square
Some quadratic functions can written as a perfect square.
x2 + 16x + 64
x2 + 26x + 169
(x + 8)2
(x + 13)2
Similarly when the coefficient of x is negative:
x2 - 6x + 9
(x - 3)2
x2 - 4x + 4
What is the relationship
between the constant term
and the coefficient of x?
(x - 2)2
The constant term is always (half the coefficient of x)2.
Completing The Square
This method enables us to write equivalent expressions for quadratics of
the form ax2 + bx. We simply half the coefficient of x to complete the
square then remember to correct for the constant term.
x2 + 4x
= (x + 2)2 ± ?
x2 + 10x
= (x + 5)2
= (x2 + 4x + 4) – 4
= (x2 + 10x + 25) - 25
= (x + 2)2 – 4
= (x + 5)2 – 25
Completing The Square
This method enables us to write equivalent expressions for quadratics of
the form ax2 + bx. We simply half the coefficient of x to complete the
square then remember to correct for the constant term.
x2 + 6x
= (x + 3)2 ± ?
x2 – 14x
= (x – 7)2
= (x2 + 6x + 9) – 9
= (x2 – 14x + 49) - 49
= (x + 3)2 – 9
= (x – 7)2 – 49
Completing The Square
x2 + 8x
= (x + 4)2 - 16
x2 - 18x
= (x - 9)2 - 81
x2 + 22x
= (x + 11)2 - 121
x2 - 2x
= (x - 1)2 - 1
x2 + 4x + 3
= (x + 2)2 - ? + 3
= (x2 + 4x + 4) - 4 + 3
= (x + 2)2 – 1
x2 + 4x + 3
Same but
simply
taking 4
off at end
= (x + 2)2+ 3 - 4
= (x + 2)2 – 1
Completing The Square
x2 + 10x + 15
x2 + 10x + 15
= (x + 5)2 - ? + 15
= (x + 5)2 + 15 - 25
= (x2 + 10x + 25) - 25 + 15
= (x + 5)2 – 10
= (x + 2)2 – 10
Same but
simply
taking 25
off at end
Completing The Square
x2 - 12x - 1
x2 + 16x - 7
= (x - 6)2 -1 - 36
= (x + 8)2 - 7 - 64
= (x - 6)2 - 37
= (x + 8)2 - 71
Completing The Square
x2 + 6x + 1
x2 + 10x + 7
= (x + 3)2 - 8
= (x + 5)2 - 18
x2 - 2x + 10
x2 - 12x - 3
= (x - 1)2 + 9
= (x - 6)2 - 39
Questions: Write the following in completed square form:
1. x2 + 8x + 10
= (x + 4)2 - 6
2. x2 - 6x + 1
= (x - 3)2 - 8
3. x2 - 2x + 2
= (x - 1)2 + 1
4. x2 + 10x + 30
= (x + 5)2 + 5
5. x2 + 6x - 5
= (x + 3)2 - 14
6. x2 - 12x - 3
= (x - 6)2 - 39
7. x2 - 4x + 5
= (x - 2)2 + 1
8. x2 - 14x - 1
= (x - 7)2 - 50
Questions 1
Completing The Square
– x2 + 10x – 15
= – (x2 – 10x) – 15
= –[(x – 5)2 - 25] – 15
= – (x – 5)2 + 25 – 15
= – (x – 5)2 + 10
Completing The Square
– x2 – 6x – 15
= – (x2 + 6x) – 15
= –[(x + 3)2 - 9] – 15
= – (x + 3)2 + 9 – 15
= – (x + 3)2 – 6
Completing The Square
21 – 12x – x2
= – (x2 + 12x) + 21
= –[(x + 6)2 - 36] + 21
= – (x + 6)2 + 36 + 21
= – (x + 6)2 + 57
Now try Exercise 1A
Page 30
Qs 1 to 16
Completing The Square
11x2 + 44x + 17
= 11(x2 + 4x) + 17
= 11[(x + 2)2 - 4] + 17
= 11(x + 2)2 - 44 + 17
= 11(x + 2)2 – 27
Completing The Square
3x2 + 30x – 1
= 3(x2 + 10x) – 1
= 3[(x + 5)2 - 25] – 1
= 3(x + 5)2 - 75 – 1
= 3(x + 5)2 – 76
Completing The Square
2x2 + 12x – 3
= 2(x2 – 6x) – 3
= 2[(x – 3)2 - 9] – 3
= 2(x – 3)2 - 18 – 3
= 2(x – 3)2 – 21
Completing The Square involving Fractions
x2 + 5x
= (x + 5/2)2 - ?
= (x2 + 2(5/2)x + (5/2)2 ) - ?
= (x2 + 5x + (25/4) )- (5/2)2
= (x + 5/2)2 - 25/4
Completing The Square involving Fractions
x2 + 7x
= (x + 7/2)2 - ?
= (x2 + 2(7/2)x + (7/2)2 ) - ?
= (x2 + 7x + (49/4) ) - (7/2)2
= (x + 7/2)2 - 49/4
Completing The Square involving Fractions
x2 – 3x
= (x – 3/2)2 – (– 3/2)2
= (x – 3/2)2 – 9/4
Completing The Square
x2 + x
x2 + 5x
= (x + 1/2)2 - (1/2)2
= (x + 5/2)2 - (5/2)2
= (x + 1/2)2 - 1/4
= (x + 5/2)2 - 25/4
Completing The Square
x2 - 7x
x2 – 9x
= (x - 7/2)2 – (-7/2)2
= (x - 9/2)2 – (-9/2)2
= (x - 7/2)2 – 49/4
= (x + 5/2)2 – 81/4
Completing The Square
x2 + ½ x
x2 – 1.4x
= (x + 1/4)2 – (1/4)2
= (x – 0.7)2 – (0.7)2
= (x + 1/4)2 – 1/16
= (x – 0.7)2 – 0.49
Completing The Square
2x2 – 6x – 3
= 2(x2 – 3x) – 3
= 2[(x – 3/2)2 - (– 3/2)2 ] – 3
= 2[(x – 3/2)2 - 9/4 ] – 3
= 2(x – 3/2)2 - 9/2 – 6/2
= 2(x – 3/2 )2 – 15/2
Completing The Square
3x2 – 21x + 11
= 3(x2 – 7x) + 11
= 3[(x – 7/2)2 - (– 7/2)2 ] + 11
= 3[(x – 7/2)2 - 49/4 ] + 11
= 3(x – 7/2)2 - 147/4 + 44/4
= 3(x – 7/2)2 – 103/4
Completing The Square
x2 + 1/3 x + 1
= (x + 1/6)2 - 1/36 + 1
= (x + 1/6)2 + 35/36
2x2 + 6x – 1/4
= 2(x + 3)2 – 1/4
= 2[(x + 3/2)2 – (3/2)2] –1/4
= 2[(x + 3/2)2 – 9/4 ] – 1/4
= 2(x + 3/2)2 – 18/4 – 1/4
= 2(x + 3/2)2 – 19/4
Completing The Square
-x2 + 2x + 9
x2 - 2.2x + 3.21
= - (x - 1)2 + 10
= (x – 1.1)2 + 2
– 1.2x + 0.26
x2 - 9x + 40
x2
= (x –
0.6)2 -
0.1
= (x – 9/2)2 – 81/2 + 40
= (x – 9/2)2 – 1/2
Completing The Square
1. – x2 + 8x – 11 = – (x – 4)2 + 5
2. – x 2 – 6x + 1
= –(x + 3)2 + 10
3. – x2 – 2x + 1
= – (x – 1)2 + 2
4. 3x2 + 12x + 7
= 3(x + 2)2 – 5
5. x2 – 5x – 7/4
= (x – 5/2)2 - 8
6. x2 + 3x – 3/4
= (x + 3/2)2 – 3
7. x2 – x + 4
= (x – ½)2 + 15/4
8. x2 – 3x + 7
= (x – 3/2)2 + 19/4
Now try Exercise 1B
Page 30