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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 +

3m + 2w = 460

2m

2m +

2m + w = 280

2m + w

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