1) The document discusses the process of completing the square to solve quadratic equations. It provides examples of factorizing quadratics and using completing the square to solve equations like x^2 - 8x - 7 = 0. 2) The document shows the step-by-step process of completing the square, which involves grouping like terms and adding and subtracting the half of the coefficient of x. 3) Examples are provided to show solving quadratics using completing the square, such as solving 3x^2 - 12x + 9 = 0.

1. Quadratics:
Completing the
square
LO 11.2 4

2. Completing the square.
Factorise the following:-
1. x2 – 3x + 9 = ( x – 3)(x – 3)
2. x2 + 8x + 16 = ( x + 4)(x + 4)
3. 2x2 – 8x + 8 = 2(x2 – 4x + 4)
= 2(x – 2)(x – 2)
□ □
4. x2 - 10x + 25 = ( x − 5 )( x − 5)

3. Completing the square
Eg1
x2 - 8 x - 7 Space out
= x2 – 8x -7
8
b
/2 : add and
= x2 - 8 x + ( 2)2 - 7 – ( 16
) subtract same
= x2 - 8 x + ( 4 )2 - 7 – 16 number
=(x–4)2 - 23

4. Completing the square
Eg1 NB coefficient of x
3x2 – 12x + 9 MUST be 1
= 3[ x2 – 4x + 3] Common factor
= 3[x2 – 4x +3 ] Space out
4 b
/2 : add and subtract
= 3[x – 4 x + ( 2 ) + 3 – ( 4)]
2 2
same number.
= 3[x – 4 x + ( 2 ) + 3 – 4]
2 2
= 3[ ( x – 2 ) 2 – 1]
Multiply both terms
=3(x–2)2 – 3 by 3

5. Completing the square.
• Do the following:-
1. x2 + 6x + 1
2. x2 – 5x + 3
3. 2x2 + 8x – 4
4. 3x2 – 9x + 2

6. Completing the square
• Solutions
1. x2 + 6x + 1
= x2 + 6x +(6/2)2 + 1 – 9
= (x + 3)2 - 8
2. x2 – 5x + 3
= x2 – 5x +( 5/2)2 + 3 – 25/4
= (x – 5/2 )2 + 3 – 61/4
= (x – 21/2)2 - 3 1/4

7. Completing the square.
• Solutions cont.
3. 2x2 + 8x – 4
= 2[ x2 + 4x –2 ]
= 2[ x2 + 4x +( 4/2)2 – 2 – 4]
= 2[ ( x + 2)2 – 6]
= 2( x + 2 )2 - 12

8. Completing the square
Solutions cont
4. 3 x 2 – 9x – 2
= 3[ x2 – 3x – 2/ 3 ]
= 3[ x2 – 3x + (3/2)2 – 2/3 – 9/4 ]
= 3[ ( x – 3/2) 2 – 8/12 – 27/12 ]
= 3[ ( x – 3/2 )2 – 35/12 ]
= 3( x – 3/2 ) 2 – 35/4 .

9. Solving for x:
x2 – 3x – 7 =0
x 2 – 3x =7
x2 - 3x + (3/2)2 = 7 + (9/4)
( x - 3/2 )2 = 28/4 + 9/4
( x – 3/2 )2 = 37/4
37
x – 3/2 =± 4
3 37 3 ± 37
x = ± =
2 4 2

10. Solving for x
1 2x2 + 8x – 4 = 0
x2 + 4x – 2 =0
x2 + 4x =2
x2 + 4x + ( 4/2)2 =2+4
( x + 2)2 = 6
x+2 =± 6
x = -2 ± 6

11. ax2 + bx + c =0
c c c
x2 + (b/a) x + − 0
= a −
a a
 b 2  b2 
x + ( /a)x +
2 b
  =  2-
 4a  /a
c
 2a   
b − 4ac
2
4a 2
( x + b/2a ) 2 = ± b 2 − 4ac
2a
x + b/2a b =
b 2 − 4ac − b ± b − 4ac 2
− ±
2a 2a 2a
x = =