Quadratic Equations Completing the Square (a) · Completing The Square This method enables us to...

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

= x2 + 6x + 9

(x + 3)2 = (x + 3)(x + 3)

= x2 + 3x + 3x + 9



(x + 5)2 = (x + 5)(x + 5)

= x2 + 5x + 5x + 25

= x2 + 10x + 25

= x2 – 4x + 4

(x – 2)2 = (x – 2)(x – 2)

= x2 – 2x – 2x + 4



= x2 + 6x + 9

= x2 – 4x + 4 (x – 2)2

(x + 5)2

(x + 1)2

(x + 3)2

= x2 + 2x + 1

= x2 + 10x + 25



= x2 + 6x + 9

= x2 – 4x + 4 (x – 2)2 (x + 5)2

(x + 1)2 (x + 3)2 = x2 + 2x + 1

= x2 + 10x + 25

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

(x + 4)2

Completing The Square

Using this idea we can factorise some quadratic functions into perfect squares.

x2 + 8x + 16 x2 + 10x + 25

(x + 5)2

x2 + 6x + 9

(x + 3)2

x2 + 14x + 49

(x + 7)2

(x – 6)2


Using this idea we can factorise some quadratic functions into perfect squares.

x2 – 12x + 36 x2 – 20x + 100

(x – 10)2 When we write

expressions in this

form it is known as

completing the

square.

x2 – 24x + 144

(x – 12)2

x2 – 2x + 1

(x – 1)2


Some quadratic functions can written as a perfect square.

x2 + 16x + 64 x2 + 26x + 169

(x + 13)2 (x + 8)2

(x - 3)2 (x - 2)2

x2 - 6x + 9 x2 - 4x + 4

Similarly when the coefficient of x is negative:

What is the relationship

between the constant term

and the coefficient of x?

The constant term is always (half the coefficient of x)2.

= (x + 5)2 = (x + 2)2

x2 + 4x x2 + 10x


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.

± ?

- 25 = (x2 + 4x + 4)

= (x + 2)2 – 4

– 4 = (x2 + 10x + 25)

= (x + 5)2 – 25

= (x – 7)2 = (x + 3)2

x2 + 6x x2 – 14x


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.

± ?

- 49 = (x2 + 6x + 9)

= (x + 3)2 – 9

– 9 = (x2 – 14x + 49)

= (x – 7)2 – 49

= (x + 11)2 = (x + 4)2

= (x - 9)2 = (x - 1)2

x2 + 8x x2 + 22x

x2 - 18x x2 - 2x


- 16 - 121

- 81 - 1

= (x + 2)2

x2 + 4x + 3 Same but

simply

taking 4

off at end

- ? + 3

= (x2 + 4x + 4) - 4 + 3

= (x + 2)2 – 1

x2 + 4x + 3

= (x + 2)2 + 3 - 4

= (x + 2)2 – 1

= (x + 5)2

x2 + 10x + 15


- ? + 15

= (x2 + 10x + 25) - 25 + 15

= (x + 5)2 – 10

x2 + 10x + 15

= (x + 5)2 + 15 - 25

= (x + 2)2 – 10

Same but

simply

taking 25

off at end

= (x - 6)2

x2 - 12x - 1


-1 - 36

= (x - 6)2 - 37

x2 + 16x - 7

= (x + 8)2 - 7 - 64

= (x + 8)2 - 71

= (x + 5)2 = (x + 3)2

= (x - 1)2 = (x - 6)2

x2 + 6x + 1 x2 + 10x + 7

x2 - 2x + 10 x2 - 12x - 3


- 8 - 18

+ 9 - 39

Questions 1

Questions: Write the following in completed square form:

1. x2 + 8x + 10

2. x2 - 6x + 1

3. x2 - 2x + 2

4. x2 + 10x + 30

5. x2 + 6x - 5

6. x2 - 12x - 3

7. x2 - 4x + 5

8. x2 - 14x - 1

= (x + 4)2 - 6

= (x - 3)2 - 8

= (x - 1)2 + 1

= (x + 5)2 + 5

= (x + 3)2 - 14

= (x - 6)2 - 39

= (x - 2)2 + 1

= (x - 7)2 - 50

= –[(x – 5)2

– x2 + 10x – 15


= – (x2 – 10x) – 15

- 25] – 15

= – (x – 5)2 + 10

= – (x – 5)2 + 25 – 15

= –[(x + 3)2

– x2 – 6x – 15


= – (x2 + 6x) – 15

- 9] – 15

= – (x + 3)2 – 6

= – (x + 3)2 + 9 – 15

= –[(x + 6)2

21 – 12x – x2


= – (x2 + 12x) + 21

- 36] + 21

= – (x + 6)2 + 57

= – (x + 6)2 + 36 + 21

Now try Exercise 1A

Page 30

Qs 1 to 16

= 11[(x + 2)2

11x2 + 44x + 17


= 11(x2 + 4x) + 17

- 4] + 17

= 11(x + 2)2 – 27

= 11(x + 2)2 - 44 + 17

= 3[(x + 5)2

3x2 + 30x – 1


= 3(x2 + 10x) – 1

- 25] – 1

= 3(x + 5)2 – 76

= 3(x + 5)2 - 75 – 1

= 2[(x – 3)2

2x2 + 12x – 3


= 2(x2 – 6x) – 3

- 9] – 3

= 2(x – 3)2 – 21

= 2(x – 3)2 - 18 – 3

= (x + 5/2)2

= (x2 + 2(5/2)x + (5/2)2 )

x2 + 5x

Completing The Square involving Fractions

- ?

- (5/2)2

- ?

= (x2 + 5x + (25/4) )

= (x + 5/2)2 - 25/4

= (x + 7/2)2

= (x2 + 2(7/2)x + (7/2)2 )

x2 + 7x

- ?

- (7/2)2

- ?

= (x2 + 7x + (49/4) )

= (x + 7/2)2 - 49/4


= (x – 3/2)2

x2 – 3x

– (– 3/2)2

= (x – 3/2)2 – 9/4


= (x + 5/2)2 = (x + 1/2)

2

x2 + x x2 + 5x


- (1/2)2 - (5/2)

2

= (x + 1/2)2 - 1/4 = (x + 5/2)

2 - 25/4

= (x - 9/2)2 = (x - 7/2)

2

x2 - 7x x2 – 9x


– (-7/2)2 – (-9/2)

2

= (x - 7/2)2 – 49/4 = (x + 5/2)

2 – 81/4

= (x – 0.7)2 = (x + 1/4)2

x2 + ½ x x2 – 1.4x


– (1/4)2 – (0.7)2

= (x + 1/4)2 – 1/16 = (x – 0.7)2 – 0.49

= 2[(x – 3/2)2

2x2 – 6x – 3


= 2(x2 – 3x) – 3

- (– 3/2)2 ] – 3

= 2(x – 3/2 )2 – 15/2

= 2(x – 3/2)2 - 9/2 – 6/2

= 2[(x – 3/2)2 - 9/4 ] – 3

= 3[(x – 7/2)2

3x2 – 21x + 11


= 3(x2 – 7x) + 11

- (– 7/2)2 ] + 11

= 3(x – 7/2)2 – 103/4

= 3(x – 7/2)2 - 147/4 + 44/4

= 3[(x – 7/2)2 - 49/4 ] + 11

= (x + 1/6)2

x2 + 1/3 x + 1


- 1/36 + 1 2x2 + 6x – 1/4

= 2[(x + 3/2)2 – (3/2)

2] –1/4

= 2(x + 3)2 – 1/4

= 2[(x + 3/2)2 – 9/4 ] – 1/4

= 2(x + 3/2)2 – 18/4 – 1/4

= (x + 1/6)2 + 35/36

= 2(x + 3/2)2 – 19/4

= - (x - 1)2

= (x – 0.6)2 = (x – 9/2)2

-x2 + 2x + 9

x2 – 1.2x + 0.26 x2 - 9x + 40


+ 10

- 0.1 – 81/2 + 40

= (x – 1.1)2

x2 - 2.2x + 3.21

+ 2

= (x – 9/2)2 – 1/2


1. – x2 + 8x – 11

2. – x 2 – 6x + 1

3. – x2 – 2x + 1

4. 3x2 + 12x + 7

5. x2 – 5x – 7/4

6. x2 + 3x – 3/4

7. x2 – x + 4

8. x2 – 3x + 7

= – (x – 4)2 + 5

= –(x + 3)2 + 10

= – (x – 1)2 + 2

= 3(x + 2)2 – 5

= (x – 5/2)2 - 8

= (x + 3/2)2 – 3

= (x – ½)2 + 15/4

= (x – 3/2)2 + 19/4

Now try Exercise 1B

Page 30

Quadratic Equations Completing the Square (a) · Completing The Square This method enables us to...

Documents

Transcript of Quadratic Equations Completing the Square (a) · Completing The Square This method enables us to...