Summarize the following files using PPT
GCSE Mathematics – Simultaneous Equations
SLIDE NUMBER 1
May 2019
© VIDLEARN® 2019
Rebecca Wigfull
1
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 2
May 2019
© VIDLEARN® 2019
2
CONSIDER…
At this point you should consider the list of session objectives and ask yourself:
How many of the session objectives am I confident with
Could I explain these objectives in relation to teaching and learning
SLIDE NUMBER 3
May 2019
© VIDLEARN® 2019
3
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 4
May 2019
© VIDLEARN® 2019
4
Simple simultaneous equations
Simultaneous Equations
SLIDE NUMBER 5
May 2019
© VIDLEARN® 2019
2x – y = 1
2x + 2y = 10
1
2
2x – y = 1
2x + 2y = 10
-3y = -9
y = 3
–
Sub y = 3 into
1
2x – 1×3 = 1
2x – 3 = 1
2x = 4
x = 2
Sub y = 3 and x = 2 into
2
2×2 + 2×3 = 10
4 + 6 = 10
10 = 10
x = 2 and y = 3
(2,3)
Check
Same Sign Subtract (SSS)
5
Simple simultaneous equations
Simultaneous Equations
SLIDE NUMBER 6
May 2019
© VIDLEARN® 2019
2x – 5y = 1
3x + 5y = 14
1
2
2x – 5y = 1
3x + 5y = 14
5x = 15
x = 3
+
Sub x = 3 into
1
2×3 – 5y = 1
6 – 5y = 1
– 5y = -5
y = 1
Sub x = 3 and y = 1 into
2
3×3 + 5×1 = 14
9 + 5 = 14
14 = 14
x = 3 and y = 1
(3,1)
Check
6
Review of main ideas from above:
Solve the following simultaneous equations
SLIDE NUMBER 7
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
3x – y = 18
3x + 6y = -3
b)
3x – 2y = 57
5x – 2y = 111
c)
2x – y = 7
4x + y = 23
x = 5 and y = -3
(5,-3)
x = 27 and y = 12
(27,12)
x = 5 and y = 3
(5,3)
7
Review of main ideas from above:
Solutions
SLIDE NUMBER 8
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
3x – y = 18
3x + 6y = -3
1
2
3x – y = 18
3x + 6y = -3
-7y = 21
y = -3
–
Sub y = -3 into
1
3x – 1x-3 = 18
3x + 3 = 18
3x = 15
x = 5
Sub x = 5 and y = -3 into
2
3×5 + 6x-3 = -3
15 – 18 = -3
-3 = -3
x = 5 and y = -3
(5,-3)
Check
Same Sign Subtract (SSS)
8
Review of main ideas from above:
Solutions
SLIDE NUMBER 9
May 2019
© VIDLEARN® 2019
CONSIDER…
b)
3x – 2y = 57
5x – 2y = 111
1
2
3x – 2y = 57
5x – 2y = 111
-2x =-54
x = 27
–
Sub x = 27 into
1
3×27 – 2y = 57
81 – 2y = 57
– 2y =-24
y = 12
Sub x = 27 and y = 12 into
2
5×27 – 2×12 = 111
135 – 24 = 111
111 = 111
x = 27 and y = 12
(27,12)
Check
Same Sign Subtract (SSS)
9
Review of main ideas from above:
Solutions
SLIDE NUMBER 10
May 2019
© VIDLEARN® 2019
CONSIDER…
1
2
2x – y = 7
4x + y = 23
6x = 30
x = 5
+
Sub x = 5 into
1
2×5 – y = 7
10 – y = 7
– y = -3
y = 3
Sub x = 5 and y = 3 into
2
4×5 + 1×3 = 23
20 + 3 = 23
23 = 23
x = 5 and y = 3
(5,3)
Check
c)
2x – y = 7
4x + y = 23
10
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 11
May 2019
© VIDLEARN® 2019
11
Engineering a match
Simultaneous Equations
SLIDE NUMBER 12
May 2019
© VIDLEARN® 2019
2x + 3y = 4
x – 2y = -5
1
2
2x + 3y = 4
2x – 4y = -10
7y = 14
y = 2
–
Sub y = 2 into
1
2x + 3×2 = 4
2x + 6 = 4
2x = -2
x = -1
Sub x = -1 and y = 2 into
2
1x-1 – 2×2 = -5
-1 – 4 = -5
-5 = -5
x = -1 and y = 2
(-1,2)
Check
Same Sign Subtract (SSS)
Multiply by 2
2x – 4y = -10
2
3
12
Engineering a match
Simultaneous Equations
SLIDE NUMBER 13
May 2019
© VIDLEARN® 2019
3x + 5y = 7
4x + 3y = 2
1
2
9x + 15y = 21
20x + 15y = 10
-11x = 11
x = – 1
–
Sub x = -1 into
1
3x-1 + 5y = 7
-3 + 5y = 7
5y = 10
y = 2
Sub x = -1 and y = 2 into
2
4x-1 + 3×2 = 2
-4 + 6 = 2
2 = 2
x = -1 and y = 2
(-1,2)
Check
Same Sign Subtract (SSS)
Multiply by 3
9x + 15y = 21
1
3
Multiply by 5
20x + 15y = 10
2
4
13
Review of main ideas from above:
Solve the following simultaneous equations
SLIDE NUMBER 14
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
2x + 3y = 8
5x – 2y = 1
b)
3x – y = 18
x + 2y = -1
c)
2x – 5y = 3
3x + 2y = 14
x = 1 and y = 2
(1,2)
x = 5 and y = -3
(5,-3)
x = 4 and y = 1
(4,1)
14
Review of main ideas from above:
Solutions
SLIDE NUMBER 15
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
2x + 3y = 8
5x – 2y = 1
x = 1 and y = 2
(1,2)
1
2
4x + 6y = 16
15x – 6y = 3
19x = 19
x = 1
+
Sub x = 1 into
1
2×1 + 3y = 8
2 + 3y = 8
3y = 6
y = 2
Sub x = 1 and y = 2 into
2
5×1 – 2×2 = 1
5 – 4 = 1
1 = 1
Check
Multiply by 2
4x + 6y = 16
1
3
Multiply by 3
15x – 6y = 3
2
4
15
SLIDE NUMBER 16
May 2019
© VIDLEARN® 2019
CONSIDER…
b)
3x – y = 18
x + 2y = -1
x = 5 and y = -3
(5,-3)
Review of main ideas from above:
Solutions
1
2
3x – y = 18
3x + 6y = -3
-7y = 21
y = -3
–
Sub y = -3 into
1
3x – 1x-3 = 18
3x + 3 = 18
3x = 15
x = 5
Sub x = 5 and y = -3 into
2
1×5 + 2x-3 = -1
5 – 6 = -1
-1 = -1
Check
Multiply by 3
3x + 6y = -3
2
3
Same Sign Subtract (SSS)
16
SLIDE NUMBER 17
May 2019
© VIDLEARN® 2019
CONSIDER…
c)
2x – 5y = 3
3x + 2y = 14
x = 4 and y = 1
(4,1)
Review of main ideas from above:
Solutions
1
2
6x – 15y = 9
6x + 4y = 28
-19y = -19
y = 1
–
Sub y = 1 into
1
2x – 5×1 = 3
2x – 5 = 3
2x = 8
x = 4
Sub x = 4 and y = 1 into
2
3×4 + 2×1 = 14
12 + 2 = 14
14 = 14
Check
Multiply by 3
6x – 15y = 9
1
3
Multiply by 2
6x + 4y = 28
2
4
Same Sign Subtract (SSS)
17
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 18
May 2019
© VIDLEARN® 2019
18
Questions in context – writing equations
Simultaneous Equations
SLIDE NUMBER 19
May 2019
© VIDLEARN® 2019
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
3m
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
3m + 2w
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
3m +
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
3m + 2w = 460
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
2m
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
2m +
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
2m + w = 280
2m + w
3 men and 2 women earn £460 per week at a company. 2 men and 1 woman earn £280. Assuming all men earn the same amount, as do the women, how much does each man and woman earn?
19
Questions in context – solving
Simultaneous Equations
SLIDE NUMBER 20
May 2019
© VIDLEARN® 2019
3m + 2w = 460 2m + w = 280
1
2
3m + 2w = 460
4m + 2w = 560
-m = -100
m = 100
–
Sub m = 100 into
1
3×100 + 2w = 460
300 + 2w = 460
2w = 160
w = 80
Sub m = 100 and w = 80 into
2
2×100 + 1×80 = 280
200 + 80 = 280
280 = 280
Men earn £100 Women earn £80
Check
Same Sign Subtract (SSS)
Multiply by 2
4m + 2w = 560
2
3
20
Review of main ideas from above:
Solve the following simultaneous equations
SLIDE NUMBER 21
May 2019
© VIDLEARN® 2019
CONSIDER…
3 lots of a number ‘x’ added to 2 lots of a number ‘y’ comes to 12. Also, 4 lots of the same number ‘x’ minus 1 lots of the number ‘y’ comes to 5. What are the values of ‘x’ and ‘y’?
3 fish minus 2 bags of chips cost £6. Also, 5 fish and 6 lots of chips costs £38. How much does one fish and a bag of chips cost?
x = 2 and y = 3
Twice one number added to three times another is 21. The difference between the number is three. Find the numbers.
Fish = £4 Chips = £3 Total £7
3 and 6
21
Review of main ideas from above:
Solutions
SLIDE NUMBER 22
May 2019
© VIDLEARN® 2019
CONSIDER…
3x + 2y = 12
4x – y = 5
x = 2 and y = 3
1
2
3x + 2y = 12
8x – 2y = 10
11x = 22
x = 2
+
Sub x = 2 into
1
3×2 + 2y = 12
6 + 2y = 12
2y = 6
y = 3
Sub x = 2 and y = 3 into
2
4×2 – 1×3 = 5
8 – 3 = 5
5 = 5
Check
Multiply by 2
8x – 2y = 10
2
3
3 lots of a number ‘x’ added to 2 lots of a number ‘y’ comes to 12. Also, 4 lots of the same number ‘x’ minus 1 lots of the number ‘y’ comes to 5. What are the values of ‘x’ and ‘y’?
22
Review of main ideas from above:
Solutions
SLIDE NUMBER 23
May 2019
© VIDLEARN® 2019
CONSIDER…
3x – 2y = 6
5x + 6y = 38
1
2
9x – 6y = 18
5x + 6y = 38
14x = 56
x = 4
+
Sub x = 4 into
1
3×4 – 2y = 6
12 – 2y = 6
– 2y = -6
y = 3
Sub x = 4 and y = 3 into
2
5×4 + 6×3 = 38
20 + 18 = 38
38 = 38
Check
Multiply by 3
9x – 6y = 18
1
3
3 fish minus 2 bags of chips costs £6. Also, 5 fish and 6 lots of chips costs £38. How much does one fish and a bag of chips cost?
Fish = £4 Chips = £3 Total £7
23
Review of main ideas from above:
Solutions
SLIDE NUMBER 24
May 2019
© VIDLEARN® 2019
CONSIDER…
2x + 3y = 21
x – y = 3
1
2
2x + 3y = 21
3x – 3y = 9
5x = 30
x = 6
+
Sub x = 6 into
1
2×6 + 3y = 21
12 + 3y = 21
3y = 9
y = 3
Sub x = 6 and y = 3 into
2
1×6 – 1×3 = 3
6 – 3 = 3
3 = 3
Check
Multiply by 3
3x – 3y = 9
2
3
Twice one number added to three times another is 21. The difference between the number is three. Find the numbers.
6 and 3
24
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 25
May 2019
© VIDLEARN® 2019
25
Simultaneous Equations
SLIDE NUMBER 26
May 2019
© VIDLEARN® 2019
y = x2 + 2x – 4
y = x + 2
x2 + 2x – 4 = x + 2
x2 + x – 6 = 0
(x+3)(x-2) = 0
x = -3 or 2
x = -3
y = x + 2
y = -3 + 2
y = -1
(-3,-1)
x = 2
y = x + 2
y = 2 + 2
y = 4
(2,4)
-1 = -32 + 2x-3 – 4
-1 = 9 -6 -4
-1 = -1
4 = 22 + 2×2 – 4
4 = 4 + 4 – 4
4 = 4
Check
Linear and non-linear simultaneous equations
26
Simultaneous Equations
SLIDE NUMBER 27
May 2019
© VIDLEARN® 2019
y = x2 – 3
y – 2x = 5
x2 – 3 = 2x + 5
x2 – 2x – 8 = 0
(x+2)(x-4) = 0
x = -2 or 4
x = -2
y = 2x + 5
y = -4 + 5
y = 1
(-2,1)
x = 4
y = 2×4 + 5
y = 8 + 5
y = 13
(4,13)
1 = -22 – 3
1 = 4 – 3
1 = 1
13 = 42 – 3
13 = 16 – 3
13 = 13
Check
y = 2x + 5
Linear and non-linear simultaneous equations
27
Simultaneous Equations
SLIDE NUMBER 28
May 2019
© VIDLEARN® 2019
x2 + y2 = 1
2x + y = 1
x2 + (1-2x)(1-2x) = 1
5×2 – 4x = 0
x = 0 or 0.8
x = 0
y = 1 – 2x
y = 1
(0,1)
x = 0.8
y = 1 – 2x
y = 1 – 1.6
y = -0.6
(0.8,-0.6)
02 + 12 = 1
1 = 1
0.82 + -0.62 = 1
0.64 + 0.36 = 1
1 = 1
Check
y = 1 – 2x
x(5x – 4) = 0
x2 + 1 – 4x + 4×2 = 1
Linear and non-linear simultaneous equations
28
Linear and non-linear simultaneous equations
Simultaneous Equations
SLIDE NUMBER 29
May 2019
© VIDLEARN® 2019
3×2 + 2xy – 6 = 0
2y – 7x = 2
3×2 + 2x(1+3.5x) – 6 = 0
10×2 + 2x – 6 = 0
y = 1 + 3.5x
(0.68,3.38)
y = 1 + 3.5x
y = -2.083587387
y = -2.08
(-0.88,-2.08)
y = 1 + 3.5x
3×2 +2x + 7×2 – 6 = 0
Factorise using quadratic formula
x = -1 ± √61
10
y = 3.383587387
x = -1 + √61
10
x = 0.68 (2dp)
x = -1 – √61
10
x = -0.88 (2dp)
x = -1 + √61
10
x = -1 – √61
10
y = 3.38 (2dp)
29
Review of main ideas from above:
Solve the following simultaneous equations, round answers to 2dp is needed.
SLIDE NUMBER 30
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
y= 2×2 – 3x + 4
y = 4x + 1
b)
3x + 2y = 7
y= x2 – 2x + 3
c)
x + y = 3
x2 + y2 = 25
(0.5, 3) and
(3 , 13)
(-0.5,4.25) and
(1 , 2)
(4.70,-1.70) and
(-1.70, 4.70)
30
Review of main ideas from above:
Solutions
SLIDE NUMBER 31
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
y= 2×2 – 3x + 4
y = 4x + 1
(0.5, 3) and
(3 , 13)
2×2 – 3x + 4 = 4x + 1
2×2 – 7x + 3 = 0
(2x-1)(x-3) = 0
x = 0.5 or 3
x = 0.5
y = 4×0.5 + 1
y = 2 + 1
y = 3
(0.5,3)
x = 3
y = 4×3 + 1
y = 12 + 1
y = 13
(3,13)
3 = 2×0.52 -3×0.5 + 4
3 = 0.5 – 1.5 + 4
3 = 3
13 = 2×32 – 3×3 + 4
13 = 18 – 9 + 4
13 = 13
Check
31
Review of main ideas from above:
Solutions
SLIDE NUMBER 32
May 2019
© VIDLEARN® 2019
CONSIDER…
x2 – 2x + 3 = 3.5 – 1.5x
x2 – 0.5x – 0.5 = 0
(x-1)(x+0.5) = 0
x = -0.5 or 1
x = -0.5
y = 3.5-1.5x-0.5
y = 3.5 + 0.75
y = 4.25
(-0.5,4.25)
x = 1
y = 3.5 – 1.5×1
y = 3.5 – 1.5
y = 2
(1,2)
4.25 = -0.52 -2x-0.5 + 3
4.25 = 0.25 + 1 + 3
4.25 = 4.25
2 = 12 – 2×1 + 3
2 = 1 – 2 + 3
2 = 2
Check
b)
3x + 2y = 7
y= x2 – 2x + 3
(-0.5,4.25) and (1 , 2)
y = 3.5 – 1.5x
y= x2 – 2x + 3
32
Review of main ideas from above:
Solutions
SLIDE NUMBER 33
May 2019
© VIDLEARN® 2019
CONSIDER…
c)
x + y = 3
x2 + y2 = 25
y = 3 – x
x2 + y2 = 25
(4.70,-1.70) and
(-1.70, 4.70)
x2 + (3-x)(3-x) = 25
2×2 – 6x – 16 = 0
y = 3 – x
(4.70,-1.70)
y = 3 – x
y = 4.701562119
y = 4.70
(-1.70,4.70)
x2 + 9 – 6x + x2 = 25
Factorise using quadratic formula
x = 3 ± √41
2
y = -1.701562119
x = 3 + √41
2
x = 4.70 (2dp)
x = 3 – √41
2
x = -1.70 (2dp)
y = -1.70 (2dp)
x2 – 3x – 8 = 0
x = 3 + √41
2
x = 3 – √41
2
33
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non-linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 34
May 2019
© VIDLEARN® 2019
34
Solving simultaneous equations graphically
Simultaneous Equations
SLIDE NUMBER 35
May 2019
© VIDLEARN® 2019
x + 3y = 9
2x + y = 8
x + 3y = 9
When
x = 0
3y = 9
y = 3
When
y = 0
x = 9
2x + y = 8
When
x = 0
y = 8
x = 4
When
y = 0
2x = 8
x
x
x
x
(3,2)
35
Solving simultaneous equations graphically
Simultaneous Equations
SLIDE NUMBER 36
May 2019
© VIDLEARN® 2019
y = x + 1
Y = 2x + 3 – x2
(-1,0)
y = x + 1
When
x = 0
y = 1
When
y = 0
x = -1
and (2,3)
| x | y | x | y |
| -5 | -32 | 1 | 4 |
| -4 | -21 | 2 | 3 |
| -3 | -12 | 3 | 0 |
| -2 | -5 | 4 | -5 |
| -1 | 0 | 5 | -12 |
| 0 | 3 |
36
Solving simultaneous equations graphically
Simultaneous Equations
SLIDE NUMBER 37
May 2019
© VIDLEARN® 2019
y = 4x + 2
x2 + y2 = 36
(1,5.9)
y = 4x + 2
When
x = 0
y = 2
When
y = 0
4x = -2
and
(-1.9,-5.7)
x = -0.5
37
Review of main ideas from above:
Solve the following simultaneous equations, graphically.
SLIDE NUMBER 38
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
y= x2 – 2x -3
2y = x + 1
b)
y = 2x -3
y = 4 – 0.5x
c)
3x + 2y = 6
x2 + y2 = 4
(-1,0) and
(3.5,2.25)
(2.8,2.6)
(0.8,1.8) and
(2,0)
38
y = 0.5
Review of main ideas from above:
Solutions
SLIDE NUMBER 39
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
y= x2 – 2x -3
2y = x + 1
(-1,0)
2y = x + 1
When
x = 0
2y = 1
When
y = 0
x = -1
and (3.5,2.25)
| x | y | x | y |
| -5 | 32 | 1 | -4 |
| -4 | 21 | 2 | -3 |
| -3 | 12 | 3 | 0 |
| -2 | 5 | 4 | 5 |
| -1 | 0 | 5 | 12 |
| 0 | -3 |
39
Review of main ideas from above:
Solutions
SLIDE NUMBER 40
May 2019
© VIDLEARN® 2019
CONSIDER…
b)
y = 2x -3
y = 4 – 0.5x
y = 2x-3
When
x = 0
y = -3
x = 1.5
When
y = 0
2x = 3
y = 4 – 0.5x
When
x = 0
y = 4
x = 8
When
y = 0
-0.5x = -4
(2.8,2.6)
x
x
x
x
40
Review of main ideas from above:
Solutions
SLIDE NUMBER 41
May 2019
© VIDLEARN® 2019
CONSIDER…
c)
3x + 2y = 6
x2 + y2 = 4
(0.8,1.8)
3x + 2y = 6
When
x = 0
2y = 6
When
y = 0
3x = 6
and (2,0)
x = 2
y = 3
41
Session Objectives
The purpose of the session is to:
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non-linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions using a graph.
SLIDE NUMBER 42
May 2019
© VIDLEARN® 2019
42
SLIDE NUMBER 43
May 2019
© VIDLEARN® 2019
CONSIDER…
End of Presentation
At this point it would be advisable to go back over the presentation. Ensure that you are fully able to deal accurately and effectively with each session objective.
You should supplement the content of this session with suitable reading, research and discussion with others.
End of presentation
Rebecca Wigfull
SLIDE NUMBER 44
May 2019
© VIDLEARN® 2019
GCSE Mathematics – Simultaneous Equations
44
GCSE Mathematics – Algebra – Linear Equations
SLIDE NUMBER 1
April 2019
© VIDLEARN® 2019
Elliott Wade
1
Session Objectives
The purpose of the session is to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear Equations
Solve Linear Equations involving multi-step skills
Solve Linear Inequalities
SLIDE NUMBER 2
April 2019
© VIDLEARN® 2019
Solve
Linear
Equation
Expression
Equation
Inequality
2
CONSIDER…
At this point you should consider the list of session objectives and ask yourself:
How many of the session objectives am I confident with
Could I explain these objectives in relation to teaching and learning
SLIDE NUMBER 3
April 2019
© VIDLEARN® 2019
3
Session Objectives
The purpose of the session is to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear equations
Solve Linear Equations involving multi-step skills
Solve Linear Inequalities
SLIDE NUMBER 4
April 2019
© VIDLEARN® 2019
Solve
Linear
Equation
Expression
Equation
Inequality
4
Differentiate between Expressions, Equations and Formula
Definitions
Put into practice
Exam style questions
Linear Equations
SLIDE NUMBER 5
April 2019
© VIDLEARN® 2019
Oracy
Linear
Equation
Expression
Equation
Inequality
5
Definitions
Equation – States that two things are equal.
For example: 6a = 12
Identity – An equation that is true for any given value substituted in.
For example: xy = yx
Formula – A mathematical rule usually containing an equals sign.
For example: Area of a circle
Expression – a group of numbers, letters and operations.
For example: 5a + 9
Linear Equations
SLIDE NUMBER 6
April 2019
© VIDLEARN® 2019
B4L is key. Time to embed into practice. Students need to be comfortable with using all four key words under this topic. And so should teachers!
Formula
Expression
Identity
Equation
Equal
6
Linear Equations
SLIDE NUMBER 7
April 2019
© VIDLEARN® 2019
Put into practice
Equation
Formula
Expression
x(x-4) = 12
x + x + x + x
V = u +at
C = d
2x +14
15t3 + 3v = 146
E=MC2
B4L is key. Time to embed into practice.
Resilience
Independence
Algebra
Equation
Deep understanding
7
Linear Equations
SLIDE NUMBER 8
April 2019
© VIDLEARN® 2019
Exam style questions
A group of numbers, letters and operations
Dan: x
Harry : x + 5
Regan 2x
Mean = Total sum ÷ number of sellers
x + x + 5 + 2x
Expression = (4x + 5) ÷ 3
B4L is key. Time to embed into practice.
Resilience
Independence
Algebra
Equation
Deep understanding
8
CONSIDER…
Consider the following question:
SLIDE NUMBER 9
April 2019
© VIDLEARN® 2019
9
Linear Equations
SLIDE NUMBER 10
April 2019
© VIDLEARN® 2019
Consider task – Answer
T = 5x + 20y
B4L is key. Time to embed into practice.
Resilience
Independence
Algebra
Equation
Deep understanding
10
Session Objectives
The purpose of the session is to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear equations
Solve linear equations involving multi-step skills
Solve linear inequalities
SLIDE NUMBER 11
April 2019
© VIDLEARN® 2019
11
Solve simple Linear Equations
Methods
One step equations
Exam style questions
Linear Equations
SLIDE NUMBER 12
April 2019
© VIDLEARN® 2019
12
Methods
Linear Equations
SLIDE NUMBER 13
April 2019
© VIDLEARN® 2019
Balance method
a + 4 = 12
a = 8
Elimination method
a + 4 = 12
To eliminate +4 from the left we move it over the equals sign.
a + 4 = 12 – 4
Whenever moving across the equals sign, invert the operation
a = 8
-4
-4
Weave solving equations into several topics.
13
One step equations
Linear Equations
SLIDE NUMBER 14
April 2019
© VIDLEARN® 2019
÷10
b = 10
x = 7
÷10
10b = 100
x3 = 343
÷2
c = 16.5
÷2
33 = 2c
-12
v = -15
-12
12 + v = – 3
Weave solving equations into several topics.
www.mathsgenie.co.uk
14
Exam Style questions
Linear Equations
SLIDE NUMBER 15
April 2019
© VIDLEARN® 2019
-5
-5
y = 7
7
+4
+4
t = 11
11
Weave solving equations into several topics.
www.mathsgenie.co.uk
15
Exam Style questions
Linear Equations
SLIDE NUMBER 16
April 2019
© VIDLEARN® 2019
÷2
÷2
y = 4
4
x3
x3
y = 18
18
Weave solving equations into several topics.
16
Task:
Complete the following one step equations:
4y = 10
10g = 37
a – 7 = -3
V + 2 = -6
= 2.7
5y = 24
SLIDE NUMBER 17
April 2019
© VIDLEARN® 2019
CONSIDER…
17
Session Objectives
Consider Task Review
SLIDE NUMBER 18
April 2019
© VIDLEARN® 2019
4y = 10
10g = 37
a – 7 = -3
V + 2 = -6
= 2.7
5y = 24
÷4
÷4
4y = 10
y = 2.5
÷10
÷10
10g = 37
g = 3.7
+7
+7
a – 7 = -3
a = 4
-2
-2
V + 2 = -6
V = – 8
x4
x4
= 2.7
w = 10.8
÷5
÷5
5y = 24
a = 4.8
18
Session Objectives
The purpose of the session is to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear equations
Solve linear equations involving multi-step skills
Solve linear inequalities
SLIDE NUMBER 19
April 2019
© VIDLEARN® 2019
19
Solve linear equations involving multi-step skills
Multi-step examples
Exam Style Questions
Common misconception
Linear Equations
SLIDE NUMBER 20
April 2019
© VIDLEARN® 2019
20
Multi-Step Examples – Two Step equations
Linear Equations
SLIDE NUMBER 21
April 2019
© VIDLEARN® 2019
4x = 12
x = 3
-3
-3
÷4
÷4
+3
+3
÷4
÷4
4x = 18
x = 4.5
Multi-Step Examples – Variables on both sides
Linear Equations
SLIDE NUMBER 22
April 2019
© VIDLEARN® 2019
x = 2
x + 3 = 5
-x
-x
-3
-3
-x
-x
-3
-3
x = 1
2x + 3 = 5
÷2
÷2
2x = 2
Multi-Step Examples – Brackets
Linear Equations
SLIDE NUMBER 23
April 2019
© VIDLEARN® 2019
-2
-2
-x
-x
Expand
2x + 2 = x +5
x + 2 = 5
x = 3
Exam Style questions
Linear Equations
SLIDE NUMBER 24
April 2019
© VIDLEARN® 2019
Perimeter = 2x + 2x + 10
Perimeter = 4x + 10
34 = 4x + 10
24 = 4x
6 = x
-10
-10
÷4
÷4
https://www.mathsgenie.co.uk/resources/64_forming-and-solving-equations.pdf
Students can sometimes discover the question themselves and attempt to answer before knowing the specific question.
Comfort with answer at the end being an uncommon format.
24
Common Misconception
Linear Equations
SLIDE NUMBER 25
April 2019
© VIDLEARN® 2019
It is not uncommon for equations to be given in a format that could leave the variable on the right instead of the left.
Consider the previous question:
34 = 4x + 10
24 = 4x
6 = x
-10
-10
÷4
÷4
Having the variable on the right does not alter the outcome of the answer.
Consider:
2 + 3 = 5 and 5 = 2 + 3
https://www.mathsgenie.co.uk/resources/64_forming-and-solving-equations.pdf
Students can sometimes discover the question themselves and attempt to answer before knowing the specific question.
Comfort with answer at the end being an uncommon format.
25
Review of main ideas from above:
Consider this angle problem from the ‘Increasingly Difficult Questions’ website.
SLIDE NUMBER 26
April 2019
© VIDLEARN® 2019
CONSIDER…
Consider Slide – Answer
Two base angles must be equal (Isosceles Triangle)
Angles in a triangle sum to 180 degrees.
An equation could be formed to show this problem…
Linear Equations
SLIDE NUMBER 27
April 2019
© VIDLEARN® 2019
3n – 10 + 3n – 10 + ? = 180
6n -20 + ? = 180
6n + ? = 200
? = 200 – 6n
Pattern
Misconception
Comfort zone
Oracy
27
Session Objectives
The purpose of the session is to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear equations
Solve linear equations involving multi-step skills
Solve linear inequalities
SLIDE NUMBER 28
April 2019
© VIDLEARN® 2019
28
Solve Linear Inequalities
Notation
Examples and approaches
Exam Style questions
Linear Equations
SLIDE NUMBER 29
April 2019
© VIDLEARN® 2019
29
Notation
Confidence in understanding notation used to show inequalities is key!
Linear Equations
SLIDE NUMBER 30
April 2019
© VIDLEARN® 2019
Essential information!
Less than
Greater than
Less than or equal to
Greater than or equal to
30
Notation continued
Linear Equations
SLIDE NUMBER 31
April 2019
© VIDLEARN® 2019
Notation
Number line
Inequaltiy
Representation
31
Examples and Approaches
Linear Equations
SLIDE NUMBER 32
April 2019
© VIDLEARN® 2019
The direction of the inequality sign is essential! So do we ever change it?… When manipulating inequalities most rules are similar to standard linear equations
Okay to do:
Simplify a side
Add/subtract from both sides
Multiply/divide both sides by a positive number.
Actions that flip the sign:
Multiply/divide both sides by a negative number.
Swapping left and right sides.
32
Tip!
More ink in the symbol means more on the diagram
Examples and Approaches – Solving Example 1
Linear Equations
SLIDE NUMBER 33
April 2019
© VIDLEARN® 2019
+10
+10
÷2
÷2
2t < 20
t < 10
Number line
Negative
Solve
Inequality
Multi Step
33
Examples and Approaches – Solving Example 2
Linear Equations
SLIDE NUMBER 34
April 2019
© VIDLEARN® 2019
-5
-5
÷-5
÷-5
– 5x ≤ – 15
x ≥ 3
Don’t forget to flip the sign here!
Number line
Negative
Solve
Inequality
Multi Step
34
Examples and Approaches – Example 3 – Number line representation
Linear Equations
SLIDE NUMBER 35
April 2019
© VIDLEARN® 2019
Can you show the following inequality on a number line?
Number line
Negative
Solve
Inequality
Multi Step
35
Task: Complete the following 5 questions from www.CorbettMaths.com
SLIDE NUMBER 36
April 2019
© VIDLEARN® 2019
CONSIDER…
1.
2.
3.
5. Show the following inequality on a number line.
4. Write the inequality shown on the number line
Session Objectives
Consider task – Answers
SLIDE NUMBER 37
April 2019
© VIDLEARN® 2019
1.
2.
3.
5. Show the following inequality on a number line.
4. Write the inequality shown on the number line
>
X ≤ -10
X is less than or equal to -4
X is less than -3
https://donsteward.blogspot.com/2012/09/linear-inequalities.html
Promotion of discussion: Is this about inequalities?
What question could you ask relating to inequalities?
What key words in the question help you model the problem?
37
Session Objectives
Exam Style Questions
From Fluency and Reasoning to Problem Solving.
SLIDE NUMBER 38
April 2019
© VIDLEARN® 2019
For what values of x is the perimeter of the square greater than the perimeter of the rectangle?
<
P of S = 12x – 8
P of R = 6x + 22
https://donsteward.blogspot.com/2012/09/linear-inequalities.html
Promotion of discussion: Is this about inequalities?
What question could you ask relating to inequalities?
What key words in the question help you model the problem?
38
Session Objectives
Exam Style Questions
From Fluency and Reasoning to Problem Solving.
SLIDE NUMBER 39
April 2019
© VIDLEARN® 2019
For what values of x is the perimeter of the square greater than the perimeter of the rectangle?
<
P of S =
P of R =
12x – 8
6x + 22
<
<
12x – 8
6x + 22
39
Session Objectives
Exam Style Questions
SLIDE NUMBER 40
April 2019
© VIDLEARN® 2019
<
<
<
12x – 8
6x + 22
-6x
-6x
+8
+8
<
6x – 8
22
÷6
÷6
6x
<
<
30
5
x
40
Session Objectives
SLIDE NUMBER 41
April 2019
© VIDLEARN® 2019
The purpose of the session was to:
Differentiate between Expressions, Equations and Formula
Solve simple Linear Equations
Solve Linear Equations involving multi-step skills
Solve Linear Inequalities
41
SLIDE NUMBER 42
April 2019
© VIDLEARN® 2019
CONSIDER…
End of Presentation
At this point it would be advisable to go back over the presentation. Ensure that you are fully able to deal accurately and effectively with each session objective.
You should supplement the content of this session with suitable reading, research and discussion with others.
End of presentation
Elliott Wade
SLIDE NUMBER 43
April 2019
© VIDLEARN® 2019
GCSE Mathematics – Algebra – Linear Equations
43
GCSE Mathematics – Quadratic Equations
SLIDE NUMBER 1
May 2019
© VIDLEARN® 2019
Rebecca Wigfull
Insert a relevant picture here to fill the whole space
1
Session Objectives
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
SLIDE NUMBER 2
May 2019
© VIDLEARN® 2019
2
CONSIDER…
At this point you should consider the list of session objectives and ask yourself:
How many of the session objectives am I confident with
Could I explain these objectives in relation to teaching and learning
SLIDE NUMBER 3
May 2019
© VIDLEARN® 2019
3
Session Objectives
SLIDE NUMBER 4
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
4
Linear Graphs
Quadratic Equations
SLIDE NUMBER 5
May 2019
© VIDLEARN® 2019
y = mx + c
5
Negative curve
Positive curve
Negative curve
Positive curve
Quadratic Graphs
Quadratic Equations
SLIDE NUMBER 6
May 2019
© VIDLEARN® 2019
y = ax2 +bx + c
6
Key Characteristics of a Quadratic Graph
Quadratic Equations
SLIDE NUMBER 7
May 2019
© VIDLEARN® 2019
Root
Root
(-2,0)
(2,0)
Turning point
(0, -4)
Positive curve
7
Key Characteristics of a Quadratic Graph
Quadratic Equations
SLIDE NUMBER 8
May 2019
© VIDLEARN® 2019
Root
Root
(-4,0)
(2,0)
Turning point
(-1, 9)
Negative curve
Y – intercept
(0, 8)
8
Key Characteristics of a Quadratic Graph
Quadratic Equations
SLIDE NUMBER 9
May 2019
© VIDLEARN® 2019
Root
Turning point
(-1, 0)
Positive curve
y – intercept
(0, 1)
Repeated root
9
Key Characteristics of a Quadratic Graph
Quadratic Equations
SLIDE NUMBER 10
May 2019
© VIDLEARN® 2019
Turning point
(3, 2)
Positive curve
y – intercept
(0, 11)
Turning point
(-1, -1)
Negative curve
(0, -2)
Y – intercept
No Roots
No Roots
10
Review of main ideas from above:
For the following 3 graphs list the key characteristic of each one.
SLIDE NUMBER 11
May 2019
© VIDLEARN® 2019
CONSIDER…
The green graph is a positive quadratic curve.
It has no roots.
The turning point is (1,2).
The intercept on the y-axis is (0,3).
The blue graph is a positive quadratic curve.
It’s roots are at (-3,0) and (3,0).
The turning point is (0,-9).
The intercept on the y-axis is (0,-9).
The red graph is a negative quadratic curve.
It has a repeated root where x = 2.
The turning point is (2,0).
The intercept on the y-axis is (0, -4).
11
Session Objectives
SLIDE NUMBER 12
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
12
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 13
May 2019
© VIDLEARN® 2019
If (x + a)(x + b) is expanded we get
(x + a)(x + b) = x2 + ax + bx + ab
= x2 + (a + b)x + ab
13
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 14
May 2019
© VIDLEARN® 2019
| Quadratic | Factorised | Rule applied |
| x2 + 5x + 6 | (x + 3)(x + 2) | A positive number in each bracket. |
| x2 – 5x + 6 | (x – 3)(x – 2) | A negative number in each bracket. |
| x2 – x – 6 | (x – 3)(x + 2) | A negative and a positive number in each bracket. |
| x2 + x – 6 | (x + 3)(x – 2) | A negative and a positive number in each bracket. |
14
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 15
May 2019
© VIDLEARN® 2019
x2 + 6x + 8
x2 + 6x + 8
(x )(x )
(x + )(x + )
x2 + 6x + 8
| a | b | total |
| 1 | 8 | 1+8 = 9 |
| 2 | 4 | 2+4 = 6 |
(x + a )(x + b )
(x + 2 )(x + 4 )
(x + 2 )(x + 4 )
= x2 + 4x + 2x + 8
= x2 + 6x + 8
15
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 16
May 2019
© VIDLEARN® 2019
x2 – 7x + 12
x2 – 7x + 12
(x )(x )
(x – )(x – )
x2 – 7x + 12
| a | b | total |
| 1 | 12 | 1+12 = 13 |
| 2 | 6 | 2+6 = 8 |
| 3 | 4 | 3+4=7 |
(x – a )(x – b )
(x – 3 )(x – 4 )
(x – 3 )(x – 4 )
= x2 – 4x – 3x + 12
= x2 – 7x + 12
16
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 17
May 2019
© VIDLEARN® 2019
x2 + 6x – 16
x2 + 6x – 16
(x )(x )
(x + )(x – )
| a | b | difference |
| 1 | 16 | 15 |
| 2 | 8 | 6 |
| 4 | 4 | 0 |
(x + 8 )(x – 2 )
(x + 8 )(x – 2 )
= x2 – 2x + 8x – 16
= x2 + 6x – 16
17
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 18
May 2019
© VIDLEARN® 2019
x2 – x – 20
x2 – x – 20
(x )(x )
(x + )(x – )
| a | b | difference |
| 1 | 20 | 19 |
| 2 | 10 | 8 |
| 4 | 5 | 1 |
(x + 4 )(x – 5 )
(x + 4)(x – 5 )
= x2 – 5x + 4x – 20
= x2 – x – 20
18
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 19
May 2019
© VIDLEARN® 2019
x2 – 36
x2 – 36
(x )(x )
(x + )(x – )
(x + 6 )(x – 6 )
(x + 6)(x – 6 )
= x2 – 6x + 6x – 36
= x2 – 36
The difference of two squares
19
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 20
May 2019
© VIDLEARN® 2019
9×2 – 36
9×2 – 36
( )( )
( + )( – )
(3x + 6 )(3x – 6 )
(3x + 6)(3x – 6 )
= 9×2 – 18x + 18x – 36
= 9×2 – 36
The difference of two squares
20
Factorising Quadratics – single bracket
Quadratic Equations
SLIDE NUMBER 21
May 2019
© VIDLEARN® 2019
x2 – 23x
x(x )
= x2 – 23x
x(x – 23 )
x(x – 23 )
2×2 + 14x
2x(x )
= 2×2 + 14x
2x(x + 7 )
2x(x + 7 )
21
Factorising Quadratics
– where the coefficient of x2 is not 1
Quadratic Equations
SLIDE NUMBER 22
May 2019
© VIDLEARN® 2019
Can you do an initial factorisation to make the coefficient of x2 1?
5×2 – 20
5(x2 – 4)
5(x + 2)(x – 2)
Difference of two squares.
22
Factorising Quadratics
– where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 23
May 2019
© VIDLEARN® 2019
3×2 + 11x + 6
(3x + )(x + )
| 3x | Total | |
| 1 | 6 | 19 |
| 2 | 3 | 11 |
| 3 | 2 | 9 |
| 6 | 1 | 9 |
(3x + 2 )(x + 3 )
23
Factorising Quadratics
– where the coefficient of x2 is not 1 (prime)
– alternative method
Quadratic Equations
SLIDE NUMBER 24
May 2019
© VIDLEARN® 2019
3×2 + 11x + 6
(3x + )(x + )
3 x 6 = 18
| a | b | total |
| 1 | 18 | 19 |
| 2 | 9 | 11 |
| 3 | 6 | 9 |
(3x + 2)(x + 3)
24
Factorising Quadratics
– where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 25
May 2019
© VIDLEARN® 2019
5×2 + 7x – 6
(5x )(x )
| a | b | |
| 1 | -6 | -29 |
| 2 | -3 | -13 |
| 3 | -2 | -7 |
| 6 | -1 | 1 |
(5x – 3)(x + 2)
| a | b | |
| -1 | 6 | 29 |
| -2 | 3 | 13 |
| -3 | 2 | 7 |
| -6 | 1 | -1 |
25
Factorising Quadratics
– where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 26
May 2019
© VIDLEARN® 2019
5×2 + 7x – 6
(5x )(x )
5 x 6 = 30
(5x – 3)(x + 2)
| a | b | Difference |
| 1 | 30 | 29 |
| 2 | 15 | 13 |
| 3 | 10 | 7 |
| 6 | 5 | 1 |
26
Factorising Quadratics
– where the coefficient of x2 is not 1 (not a prime)
Quadratic Equations
SLIDE NUMBER 27
May 2019
© VIDLEARN® 2019
8×2 – 2x – 15
(2x )(4x )
| a | b | |
| 1 | -15 | -119 |
| 3 | -5 | -37 |
| 5 | -3 | -19 |
| 15 | -1 | 7 |
| a | b | |
| -1 | 15 | 119 |
| -3 | 5 | 37 |
| -5 | 3 | 19 |
| -15 | 1 | -7 |
(8x )(x )
27
Factorising Quadratics
– where the coefficient of x2 is not 1 ( not a prime)
Quadratic Equations
SLIDE NUMBER 28
May 2019
© VIDLEARN® 2019
8×2 – 2x – 15
(2x )(4x )
| a | b | |
| 1 | -15 | -26 |
| 3 | -5 | 2 |
| 5 | -3 | 14 |
| 15 | -1 | 58 |
| a | b | |
| -1 | 15 | 26 |
| -3 | 5 | -2 |
| -5 | 3 | -14 |
| -15 | 1 | -58 |
(2x -3)(4x + 5)
28
Factorising Quadratics
where the coefficient of x2 is not 1 ( not a prime)
alternative method
Quadratic Equations
SLIDE NUMBER 29
May 2019
© VIDLEARN® 2019
8×2 – 2x – 15
| a | b | |
| 1 | 120 | 119 |
| 2 | 60 | 58 |
| 3 | 40 | 37 |
| 4 | 30 | 26 |
| a | b | |
| 5 | 24 | 19 |
| 6 | 20 | 14 |
| 8 | 15 | 7 |
| 10 | 12 | 2 |
8×15 = 120
8×2 + 10x – 12x – 15
2x(4x + 5) – 3(4x + 5)
(2x-3)(4x + 5)
29
Review of main ideas from above:
Factorise each of the following questions
SLIDE NUMBER 30
May 2019
© VIDLEARN® 2019
CONSIDER…
x2 – 12x + 35
x2 – 17x + 30
x2 + 9x + 20
x2 + 19x + 18
x2 + 2x – 48
x2 – 1
3×2 – 15x
4×2 – 64
(x-7)(x-5)
(x-15)(x-2)
(x+4)(x+5)
(x+18)(x+1)
(x+8)(x-6)
(x+1)(x-1)
3x(x-5)
(2x+8)(2x-8)
2×2 + 5x +3
5×2 – 38x +21
6×2 +17x + 12
9×2 +9x – 10
(2x+3)(x+1)
(5x-3)(x-7)
(3x+4)(2x+3)
(3x-2)(3x+5)
30
Session Objectives
SLIDE NUMBER 31
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
31
Solving a quadratic equation
Quadratic Equations
SLIDE NUMBER 32
May 2019
© VIDLEARN® 2019
Root
Root
(0,-2)
(0, 2)
x2 – 4
(x-2)(x+2)= 0
x = -2 or 2
32
Solving a Quadratic Equation
Quadratic Equations
SLIDE NUMBER 33
May 2019
© VIDLEARN® 2019
Root
Root
(-4,0)
(2, 0)
-(x+4)(x-2)= 0
x = -4 or 2
-x2 – 2x + 8
-(x2 +2x – 8)= 0
33
Solving Quadratic Equations
Quadratic Equations
SLIDE NUMBER 34
May 2019
© VIDLEARN® 2019
x2 + 6x – 16 = 0
(x + 8 )(x – 2 ) = 0
(x + 8 )= 0 and (x – 2 ) = 0
x + 8 = 0
x = -8
and
x – 2 = 0
x = 2
The roots are (2,0) and (-8,0)
34
Solving Quadratic Equations
– where the coefficient of x2 is not 1
Quadratic Equations
SLIDE NUMBER 35
May 2019
© VIDLEARN® 2019
8×2 – 2x – 15 = 0
(2x -3)(4x + 5) = 0
(2x – 3)= 0 and (4x + 5) = 0
2x – 3 = 0
2x = 3
x = 1.5
and
4x + 5 = 0
4x = -5
x = -1.25
The roots are (1.5,0) and (-1.25,0)
35
Review of main ideas from above:
Solve each of the following questions you previously factorised
SLIDE NUMBER 36
May 2019
© VIDLEARN® 2019
CONSIDER…
x2 – 12x + 35=0
x2 – 17x + 30=0
x2 + 9x + 20=0
x2 + 19x + 18=0
x2 + 2x – 48=0
x2 – 1=0
3×2 – 15x=0
4×2 – 64=0
x = 5 or 7
x = 2 or 15
x = -2 or -15
x = -18 or -1
x = -8 or 6
x = 1 or -1
x = 0 or 5
x = 4 or -4
2×2 + 5x +3=0
5×2 – 38x +21=0
6×2 +17x + 12=0
9×2 +9x – 10=0
x = -1.5 or -1
x = 0.6 or 7
x = -1.3 or -1.5
x = -1.6 or 0.6
.
.
.
36
Session Objectives
SLIDE NUMBER 37
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
37
Completing the Square
Quadratic Equations
SLIDE NUMBER 38
May 2019
© VIDLEARN® 2019
Minimum Point
Minimum point (-3, -21)
Minimum value is -21
X2 + 6x – 12
Turning point
38
x2 + 6x – 12 = 0
(x + 3)2 – 9 – 12 = 0
(x + 3)2 – 21 = 0
(x + 3)2 = 21
x + 3 = ± 21
x = ± 21 – 3
x = 1.58 or x = -7.58 (2dp)
Solving using Completing the Square
Quadratic Equations
SLIDE NUMBER 39
May 2019
© VIDLEARN® 2019
Roots (1.58,0) and (-7.58,0)
Half the coefficient of x
Take away the square of half the coefficient of x
Completed square form
Minimum point (-3, -21)
Minimum value is -21
√
√
39
Completing the Square
Quadratic Equations
SLIDE NUMBER 40
May 2019
© VIDLEARN® 2019
Turning point
(-1, 9)
Maximum point
-x2 – 2x + 8
Maximum point (-1, 9)
Maximum value is 9
40
-x2 – 2x + 8 = 0
-(x2 + 2x – 8) = 0
x2 + 2x – 8 = 0
(x + 1)2 – 1 – 8 = 0
(x + 1)2 – 9 = 0
(x + 1)2 = 9
x + 1 = ± 9
x = ± 3 – 1
x = -4 or x = 2
√
Solving using Completing the Square
Quadratic Equations
SLIDE NUMBER 41
May 2019
© VIDLEARN® 2019
Roots (-4,0) and (2,0)
Completed square form
Maximum point (-1, 9)
Maximum value is 9
41
4×2 – 8x + 1 = 0
4(x2 – 2x + 0.25) = 0
x2 – 2x + 0.25 = 0
(x – 1)2 -1 + 0.25 = 0
(x – 1)2 – 0.75 = 0
(x – 1)2 = 0.75
x – 1 = ± 0.75
x = 1 ± 0.75
x = 1.87 (2dp) or x = 0.13 (2dp)
√
√
Solving using Completing the Square
Quadratic Equations
SLIDE NUMBER 42
May 2019
© VIDLEARN® 2019
Roots (1.87,0) and (0.13,0)
Minimum point (1, -0.75)
Minimum value is -0.75
42
2×2 = 8x +1 1
2×2 – 8x -11 = 0
2(x2 – 4x – 5.5) = 0
x2 – 4x – 5.5 = 0
(x – 2)2 -4 -5.5 = 0
(x – 2)2 -9.5 = 0
(x – 2)2 = 9.5
x – 2 = ± 9.5
x = 2 ± 9.5
x = 5.08 (2dp) or x = -1.08 (2dp)
√
√
Solving using Completing the Square
Quadratic Equations
SLIDE NUMBER 43
May 2019
© VIDLEARN® 2019
Minimum point (2, -9.5)
Minimum value is -9.5
43
Review of main ideas from above:
Use the completing the square to find the turning point and to solve the following quadratics. (Give answers to 2dp where necessary)
SLIDE NUMBER 44
May 2019
© VIDLEARN® 2019
CONSIDER…
| 1. x2 – 8x = -13 | 2. 2×2 + 10x + 5 = 0 |
| 3. x2 – 3x – 11 = 0 | 4. 3×2 + 2 = 9x |
| x = 5.73 (2dp) or x = 2.27 (2dp) Turning point (4,-3) | x = -0.56 (2dp) or x = -4.44 (2dp) Turning point (-2.5,-3.75) |
| x = 5.14 (2dp) or x = -2.14 (2dp) Turning point (1.5,-13.25) | x = 2.76 (2dp) or x = 0.24 (2dp) Turning point (1.5-1.58) |
44
Session Objectives
SLIDE NUMBER 45
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
45
The quadratic formula
Quadratic Equations
SLIDE NUMBER 46
May 2019
© VIDLEARN® 2019
By completing the square on the quadratic equation
ax2 + bx + c = 0
a quadratic formula can be found which will solve most quadratic equations which do not factorise.
This is an alternative method of solving equations other than by completing the square.
The quadratic formula for solving quadratic equations is
x = -b ± b2 – 4ac
2a
√
46
The quadratic formula
Quadratic Equations
SLIDE NUMBER 47
May 2019
© VIDLEARN® 2019
x = -b ± b2 – 4ac
2a
√
x2 – 2x – 6 = 0
ax2 + bx + c = 0
then a = 1, b = -2 and c = -6
x = -(-2) ± (-2)2 – 4(1)(-6)
2(1)
√
x = 2 ± 4 – -24
2
√
x = 2 ± 28
2
√
x = 2 ± 2 7
2
√
x = 1 ± 7
√
x = 3.645751311 or
x = -1.645751311
x = 3.65 (2dp) or
x = -1.65 (2dp)
47
The quadratic formula
Quadratic Equations
SLIDE NUMBER 48
May 2019
© VIDLEARN® 2019
x = -b ± b2 – 4ac
2a
√
3 x2 + 4x – 2 = 0
ax2 + bx + c = 0
then a = 3, b = 4 and c = -2
x = -(4) ± (4)2 – 4(3)(-2)
2(3)
√
x = -4 ± 16 – -24
6
√
x = -4± 40
6
√
x = -4 ±2 10
6
√
x = 0.3874258867 or
x = -1.72075922
x = -2 ± 10
3
√
x = 0.39 (2dp) or
x = -1.72 (2dp)
48
Review of main ideas from above:
Use the quadratic formula to solve the following quadratics
SLIDE NUMBER 49
May 2019
© VIDLEARN® 2019
CONSIDER…
| 1. 3×2 + 6x – 7 = 0 | 2. x2 + 3x + 1 = 0 |
| 3. 8×2 – 6x + 1 = 0 | 4. 4×2 + 7x – 6 = 0 |
| x = 0.83 (2dp) or x = -2.83 (2dp) | 2. x = -0.38 (2dp) or x = -2.62 (2dp) |
| 3. x = 0.5 or x = 0.25 | x = 0.63 (2dp) or x = -2.38 (2dp) |
49
Session Objectives
SLIDE NUMBER 50
May 2019
© VIDLEARN® 2019
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
50
Solving Quadratic inequalities
Quadratic Equations
SLIDE NUMBER 51
May 2019
© VIDLEARN® 2019
x2 + 2x -15
(x+5)(x-3)
Roots: (-5,0)and (3,0)
If …
x2 + 2x -15 ≥ 0
above x-axis
x ≥ 3 and x ≤ -5
51
Solving Quadratic inequalities
Quadratic Equations
SLIDE NUMBER 52
May 2019
© VIDLEARN® 2019
x2 + 2x -15
(x+5)(x-3)
Roots: (-5,0)and (3,0)
If …
x2 + 2x -15 ≤ 0
below x-axis
-5 ≤ x ≤ 3
52
Solving Quadratic inequalities
Quadratic Equations
SLIDE NUMBER 53
May 2019
© VIDLEARN® 2019
x2 + 12x + 82 > 22 – 4x
x2 + 16x + 60 > 0
(x + 6)(x + 10)
Roots: (-6,0)and (-10,0)
x2 + 16x + 60 > 0
above x-axis
x > -6 and x < -10
