1. 07/12/2018 Starter L.O. To be able to Solve a quadratic by factorising
Solve a quadratic by using the formula
Solve a quadratic by completing the square
Key Words
quadratic, factor, coefficient, discriminant, significant figures, surd form, formula
Starter

2. CONTENTS – Click to go to…
07/12/2018
CONTENTS – Click to go to…
BY FACTORISING
BY USING THE FORMULA
BY COMPLETING THE SQUARE
PROBLEM SOLVING WITH QUADRATICS
THE DISCRIMINANT

3. How to solve a quadratic
07/12/2018
How to solve a quadratic
Non calculator paper…
Calculator paper…
First try to FACTORISE
Does it say give your answers to 3s.f.? NO
YES
If you cannot factorise, use the FORMULA or COMPLETE THE SQUARE and leave the answers in surd form
See if you can FACTORISE first, as that is usually quickest. But remember you can always use your calculator to apply the FORMULA
Use the FORMULA, put it in your calculator and round

4. 07/12/2018
BY FACTORISING

5. 07/12/2018
Skills check – Can you answer all of the questions below?
Expand 4𝑥(𝑥 – 3)
Factorise 4𝑥2 – 12𝑥
Expand and simplify (𝑥 + 6) (𝑥 + 3)
Factorise 𝑥 2 −11𝑥 +18

6. 07/12/2018
Task – Spot the errors. How many errors can you find? If you think you found them all, then CORRECT THEM!
a) Expand and simplify (𝑥 + 4) (𝑥 – 5)
𝟐𝒙 − 𝟓𝒙 + 𝟒𝒙 + 𝟐𝟎
= 𝟏𝟏𝒙 + 𝟐𝟎
b) Factorise the following quadratic expression...
𝑥² + 11𝑥 + 18
(𝒙 + 𝟔) (𝒙 + 𝟑)

7. 12/7/2018
Factorize the given expression by finding two integers that add together to give the coefficient of x and multiply together to give the constant.
It may be a good idea to practice adding and multiplying negative numbers before attempting this activity. Use slide 31 in N1.2 Calculating with integers to do this if required.
The lower of the two hidden integers will be given first in each case.

8. 12/7/2018
Select a quadratic expression and ask a volunteer to find its corresponding factorization.

9. What is the meaning of the word “hence” in part b)?
07/12/2018
Discussion - Below is an exam question. Discuss each bullet point in pairs and be ready to give feedback
What is the difference between the algebra given to you in parts a) and b)?
What type of answer should you get for part a)? What about part b)? How do you know?
What is the meaning of the word “hence” in part b)?
a) Factorise the following quadratic expression
𝑥² + 10𝑥 + 21
b) Hence solve the equation 𝑥² + 10𝑥 + 21 = 0

10. Can you think of any values for a and b that would make this true…?
07/12/2018
Discussion
𝑎 × 𝑏 = 0
Can you think of any values for a and b that would make this true…?
How many pairs can you come up with?
Most ideas wins a prize!
You have 30 seconds…

11. 07/12/2018
Solving by factorising
Arrange so that all terms are on one side, so the equation is in the form
𝑎 𝑥 2 +𝑏𝑥+𝑐=0
FACTORISE
Make the expression inside each bracket =𝟎
SOLVE BOTH to give two possible values for 𝑥

12. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE by factorising
𝑥² + 7𝑥 + 6=0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑥+6 𝑥+1 =0
FACTORISE
Make each bracket = 0
𝑥+6=0
𝑥+1=0
SOLVE BOTH equations
𝑥=−6
𝑥=−1

13. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE by factorising
𝑥2 – 5𝑥 = –4
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑥2 – 5𝑥+4=0
𝑥−4 𝑥−1 =0
FACTORISE
Make each bracket = 0
𝑥−4=0
𝑥−1=0
SOLVE BOTH equations
𝑥=4
𝑥=1

14. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE by factorising
𝑥2 = 3𝑥
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑥2 –3𝑥=0
𝑥 𝑥−3 =0
FACTORISE
Make each bracket = 0
𝑥=0
𝑥−3=0
SOLVE BOTH equations
𝑥=0
𝑥=3

15. Example – SOLVE by factorising 𝑥2 −64=0
07/12/2018
Example – SOLVE by factorising
𝑥2 −64=0
What special case of quadratic factorising is this…?
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
(𝑥+8) 𝑥−8 =0
FACTORISE
Make each bracket = 0
𝑥+8=0
𝑥−8=0
SOLVE BOTH equations
𝑥=−8
𝑥=8

16. 07/12/2018
Example – SOLVE by factorising
𝑥² + 8𝑥 + 12=0

17. 07/12/2018
Example – SOLVE by factorising
𝑥² − 7𝑥 + 12=0

18. 07/12/2018
Example – SOLVE by factorising
𝑥² + 3𝑥 – 28=0

19. 07/12/2018
It may be a good idea to practice adding and multiplying negative numbers before attempting this activity. See the activity in N1.2 Calculating with integers.

20. Task – SOLVE by factorising
07/12/2018
Task – SOLVE by factorising
(a) 5x2 = 10x
x = 0 or 2
(b) 4x2 - 6x = 0
x = 0 or 1½
(c) x2 + 3x + 2= 0
x = -1 or -2
(d) 4x2 - 9= 0
x = +/- 1½
(e) 2t2 - 9t - 5= 0
x = -½ or 5
(f) 16x2 = 100
x = +/- 2½
(g) 5x2 = - 4x2 + 1
x = +/- 1/3
(h) 2(x2 + 5x) = -12
x = -2 or -3
(i) 12x2 - 13x + 3= 0
x = ¾ or 1/3

21. x² - 6x + 8 x² + 7x + 6 x² + x - 12 x² - 6x + 9 x² - 5x - 36
07/12/2018
Task
x² - 6x + 8
x² + 7x + 6
x² + x - 12
x² - 6x + 9
x² - 5x - 36
Hence solve the equations above

22. 07/12/2018
Task – SOLVE by factorising
x² + 7x + 10 = 0
x² + 11x + 30 = 0
x² + 4x + 3 = 0
x² + 10x + 21 = 0
x² + 7x + 12 = 0
x² + 2x + 1 = 0
x² + 5x + 6 = 0
x² + x – 30 = 0
x² – 12x + 35 = 0
x² – 4 = 0
x² + 5x – 6 = 0
x² – 12x + 35 = 0
x² – 2x – 15 = 0
x² – 5x – 36 = 0
x² – 64 = 0
4x² = 0
x² – 2.5x = 0
x² – x = 6
x² – 10 = 3x
x² – 36 = 2x - 1

23. 07/12/2018
Task – SOLUTIONS
x = -5 or -2
x = -6 or -5
x = -3 or -1
x = -7 or -3
x = -4 or -3
x = -1 or -1
x = -3 or -2
x = -6 or 5
x = 7 or 5
x = 2 or -2
x = -6 or 1
x = 7 or 5
x = 5 or -3
x = 9 or -4
x = 8 or -8
x = 5.5 or -5.5
x = 1 or 1.5
x = 3 or -2
x = 5 or -2

24. 07/12/2018
Task – SOLVE by factorising
x² + 9x = 0
x² + 9x = 0
x² + 11x = 0
x² + 10x = 0
x² + 7x = 0
x² + 3x – 18 = 0
x² - 3x – 18 = 0
x² + 3x – 28 = 0
x² - x – 12 = 0
x² + 2x – 24 = 0
x² - 8x + 7 = 0
x² - 12x + 11 = 0
x² - 5x + 4 = 0
x² - 5x + 6 = 0
x² - 9x + 8 = 0

25. 07/12/2018
Use this activity to demonstrate graphically the solutions to quadratic equations that factorize and do not factorize.
Reveal the coordinates of the points that intersect the x-axis and relate these to the solution of the equation.
Establish that without the use of a computer or a graphics calculator this is not an accurate method of solving the equation. For a more accurate solution we need to use an algebraic method. It is useful, however, for demonstrating the nature of the solutions, for example where there is only one solution, or no solutions.
The use of graphs to solve equations in covered in more detail in A9.3

26. 07/12/2018
BY USING THE FORMULA

27. 07/12/2018 Solving by using the quadratic formula
Arrange so that all terms are on one side, so the equation is in the form
𝑎 𝑥 2 +𝑏𝑥+𝑐=0
Write down 𝑎 = , 𝑏 = , 𝑐 =
Write out the quadratic formula…
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Sub in your values of 𝑎, 𝑏 and 𝑐, remembering to put BRACKETS around your values, particularly if subbing in a NEGATIVE
Work out the two solutions, either on a calculator, or leaving in surd form if you don’t have one.

28. 07/12/2018
Task – These quadratics need to be solved using the formula. Write down the values of 𝑎, 𝑏 and 𝑐 in each case…

29. 07/12/2018
The quadratic formula
MEMORISE THIS!!!!!
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

30. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
𝑥² − 7𝑥 +8=0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎=1 , 𝑏=(−7), 𝑐=8
𝑎= , 𝑏= , 𝑐= ?
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Write out the formula
𝑥= −(−7)± (−7) 2 −4(1)(8) 2(1)
SUBSTITUTE
𝑥= −32 2
𝑥= 7− 49−32 2
CALCULATE
FULL answers 𝑥= 𝑥=
𝑥=5.56 (3 𝑠.𝑓.)
𝑥=1.44 (3 𝑠.𝑓.)
ROUNDED answers

31. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
2𝑥²+5𝑥 −1=0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎=2 , 𝑏=5, 𝑐=(−1)
𝑎= , 𝑏= , 𝑐= ?
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Write out the formula
𝑥= −(5)± (5) 2 −4(2)(−1) 2(2)
SUBSTITUTE
𝑥= −
𝑥= −5−
CALCULATE
FULL answers 𝑥=
𝑥=−
𝑥=0.186 (3 𝑠.𝑓.)
𝑥=−2.69 (3 𝑠.𝑓.)
ROUNDED answers

32. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
𝑥²−5𝑥+2=0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑐= ?
Write out the formula
SUBSTITUTE
CALCULATE
FULL answers
ROUNDED answers

33. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
2𝑥²−1=−6𝑥
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑐= ?
Write out the formula
SUBSTITUTE
CALCULATE
FULL answers
ROUNDED answers

34. Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
07/12/2018
EXTENSION Example – SOLVE, giving answers to 3 s.f.
2𝑥 2 =𝑥 𝑥−14 −5
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑐= ?
Write out the formula
SUBSTITUTE
CALCULATE
FULL answers
ROUNDED answers

35. 07/12/2018
Task – On your mini whiteboard…
SOLVE the following to 2 d.p.

36. 07/12/2018
Task – SOLVE, giving your answers to 4 d.p. 1) 5)
V= OR v=
f= OR f= 2) 6)
h= OR h=1.8170
h= OR h= 3) 7)
q= OR q=
x= OR x= 4) 8)
f= OR f=7.5311
y= OR y=0.9361

37. Task – SOLVE, giving your answers to 2 d.p. 1) 𝑥² – 6𝑥 – 2 = 0
07/12/2018
Task – SOLVE, giving your answers to 2 d.p.
1) 𝑥² – 6𝑥 – 2 = 0
2) 𝑥² + 4𝑥 + 1 = 0
3) 𝑥² + 10𝑥 – 12 = 0
4) 𝑥² – 2𝑥 – 7 = 0
5) 𝑥² + 8𝑥 + 5 = 0
6) 𝑥² – 9𝑥 + 6 = 0
7) 𝑥² + 𝑥 – 8 = 0
1) 𝑥2 − 2𝑥=6
2) 𝑥2 + 5 =−7𝑥
3) 2𝑥2= 3𝑥+ 6
4) 5𝑥 + 1=−3 𝑥 2
5) 𝑥2 + 8𝑥 = 10
6) 5𝑥 – 1 – 2𝑥2 = 0
7) 3 − 2𝑥 = 4𝑥2
x = 6.32 or – 0.32
x = 3.65 or – 1.65
x = or – 3.73
x = or – 6.19
x = 2.68 or – 1.14
x = 1.08 or – 11.08
x = or – 1.43
x = 3.83 or – 1.83
x = 1.10 or – 9.10
x = or – 7.32
x = 0.22 or 2.28
x = 8.27 or 0.73
x = 0.65 or – 1.15
x = 2.37 or – 3.37

38. BY COMPLETING THE SQUARE
07/12/2018
BY COMPLETING THE SQUARE

39. 07/12/2018
Perfect squares
Some quadratic expressions can be written as perfect squares. For example,
x2 + 2x + 1 = (x + 1)2
x2 – 2x + 1 = (x – 1)2
x2 + 4x + 4 = (x + 2)2
x2 – 4x + 4 = (x – 2)2
x2 + 6x + 9 = (x + 3)2
x2 – 6x + 9 = (x – 3)2
In general,
x2 + 2ax + a2 = (x + a)2
and
x2 – 2ax + a2 = (x – a)2
How could the quadratic expression x2 + 6x be made into a perfect square?
We could add 9 to it.

40. 07/12/2018
Adding 9 to the expression x2 + 6x to make it into a perfect square is called completing the square.
x2 + 6x = x2 + 6x + 9 – 9
We can write
If we add 9 we then have to subtract 9 so that both sides are still equal.
By writing x2 + 6x + 9 we have completed the square and so we can write this as
x2 + 6x = (x + 3)2 – 9
x2 + bx = x – b 2
In general,

41. 07/12/2018 Examples Complete the square for x2 – 10x.
Compare this expression to (x – 5)2 = x2 – 10x + 25
x2 – 10x = x2 – 10x + 25 – 25
= (x – 5)2 – 25
Complete the square for x2 – 3x.
Compare this expression to (x – 1.5)2 = x2 – 3x
x2 – 3x = x2 – 3x – 2.25
= (x + 1.5)2 – 2.25

42. 07/12/2018

43. 07/12/2018
Task – Write in the form (x + p)2 + q
x² + 4x
x² + 10x
x² + 12x
x² + 24x
x² + 22x
x² – 14x
x² – 2x
x² – 16x
x² – 18x
x² – 3x
x² + 20x + 96
x² + 16x + 63
x² + 22x + 132
x² + 36x + 298
x² + 18x + 76
y² – 14y + 48
y² – 20y + 98
y² – 2y – 5
y² + 41y + 420
y² + 56y + 663

44. 07/12/2018
Task – SOLUTIONS
(x + 2)² – 4
(x + 5)² – 25
(x + 6)² – 36
(x + 12)² – 144
(x + 11)² – 121
(x – 11)² – 121
(x – 1)² – 1
(x – 8)² – 64
(x – 9)² – 81
(x – 1.5)² – 2.25
(x + 10)² – 4
(x + 8)² – 1
(x + 11)² + 10
(x + 18)² – 26
(x + 9)² – 5
(y – 7)² – 1
(y – 10)² – 2
(y – 1)² – 6
(x )² +0.25
(x + 28)² – 121

45. How can we complete the square for
07/12/2018
Completing the square
How can we complete the square for
x2 + 8x + 9?
Look at the coefficient of x.
This is 8 so compare the expression to (x + 4)2 = x2 + 8x + 16
x2 + 8x + 9 = x2 + 8x + 16 –
= (x + 4)2 – 7
x2 + bx + c = x – c b 2
In general,

46. 07/12/2018 Examples Complete the square for x2 + 12x – 5.
Compare this expression to (x + 6)2 = x2 + 12x + 36
x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5
= (x2 + 6) – 41
Complete the square for x2 – 5x + 16
Compare this expression to (x – 2.5)2 = x2 – 5x
x2 – 5x + 16 = x2 – 5x –
= (x2 – 2.5)

47. 07/12/2018

48. 1. x2 + 8x + 10 = (x + 4)2 - 6 2. x2 - 6x + 1 = (x - 3)2 - 8
07/12/2018
Task – Write each of the following in “completed square” form
1. x2 + 8x + 10
= (x + 4)2 - 6
2. x2 - 6x + 1
= (x - 3)2 - 8
3. x2 - 2x + 2
= (x - 1)2 + 1
4. x2 + 10x + 30
= (x + 5)2 + 5
5. x2 + 6x - 5
= (x + 3)2 - 14
6. x2 - 12x - 3
= (x - 6)2 - 39
7. x2 - x + 4
= (x - ½)
8. x2 - 3x
= (x - 1.5)2 + 5

49. 07/12/2018
Solving by completing the square
Arrange so that all terms are on one side, so the equation is in the form
𝑎 𝑥 2 +𝑏𝑥+𝑐=0
Write down 𝑎 = , 𝑏 = , 𝒃 𝟐 = , 𝑐 =
Complete the square using this formula
Sub in your values of 𝑎, 𝑏 and 𝑐
Now rearrange using inverse operations to get your two solutions
REMEMBER when you square root you get TWO possible solutions, so don’t forget to write the “±”
x2 + bx + c = x – c b 2

50. Example – SOLVE, giving answers to 3 s.f. 𝑥²+8𝑥 +5=0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
𝑥²+8𝑥 +5=0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎=1 , 𝑏=8, 𝑏 2 =4, 𝑐=5
𝑎= , 𝑏= , 𝑏 2 = , 𝑐= ?
𝑥+ 𝑏 − 𝑏 𝑐=0
Re-write using this method
SUBSTITUTE
𝑥+4 2 − =0
SOLVE
𝑥+4 2 −16+5=0
𝑥+4 2 −11=0
𝑥+4 2 =11
𝑥+4 =± 11
𝑥 =−4 ± 11
FULL answers
𝑥=−
𝑥=−
ROUNDED answers
𝑥=−0.683 (3 𝑠.𝑓.)
𝑥=−7.32 (3 𝑠.𝑓.)

51. Example – SOLVE, giving answers to 3 s.f. x2 – 4x – 3 = 0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
x2 – 4x – 3 = 0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑏 2 = , 𝑐= ?
Re-write using this method
SUBSTITUTE
SOLVE
FULL answers
ROUNDED answers

52. Example – SOLVE, giving answers to 3 s.f. x2 + 4x − 6 = 0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
x2 + 4x − 6 = 0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑏 2 = , 𝑐= ?
Re-write using this method
SUBSTITUTE
SOLVE
FULL answers
ROUNDED answers

53. Example – SOLVE, giving answers to 3 s.f. x2 + 6x + 3 = 0
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
x2 + 6x + 3 = 0
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑏 2 = , 𝑐= ?
Re-write using this method
SUBSTITUTE
SOLVE
FULL answers
ROUNDED answers

54. Example – SOLVE, giving answers to 3 s.f. x2 + 8x + 6 = − 7
07/12/2018
Example – SOLVE, giving answers to 3 s.f.
x2 + 8x + 6 = − 7
Put in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎= , 𝑏= , 𝑏 2 = , 𝑐= ?
Re-write using this method
SUBSTITUTE
SOLVE
FULL answers
ROUNDED answers

55. 07/12/2018
Task – Solve by completing the square
1. x2 + 6x + 1 = 0
x = and
2. x2 - 8x + 3 = 0
x = 7.61 and 0.39
3. x2 - 4x - 7 = 3
x = 5.74 and
4. x2 + 3x + 1= 0
x = and

56. Complete the square for 2x2 + 8x + 3.
07/12/2018
When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square.
Complete the square for 2x2 + 8x + 3.
Start by factorizing the first two terms by dividing by 2,
2x2 + 8x + 3 = 2(x2 + 4x) + 3
By completing the square, x2 + 4x = (x + 2)2 – 4 so,
2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3
= 2(x + 2)2 – 8 + 3
= 2(x + 2)2 – 5

57. Complete the square for 5 + 6x – 3x2.
07/12/2018
Complete the square for 5 + 6x – 3x2.
Start by factorising the terms containing x’s by –3.
5 + 6x – 3x2 = 5 – 3(–2x + x2)
5 + 6x – 3x2 = 5 – 3(x2 – 2x)
By completing the square, x2 – 2x = (x – 1)2 – 1 so,
5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1)
= 5 – 3(x – 1)2 + 3
= 8 – 3(x – 1)2

58. 12/7/2018

59. 07/12/2018 Proof of the quadratic formula by completing the square…
 by a
Complete
the square

60. PROBLEM SOLVING WITH QUADRATICS
07/12/2018
PROBLEM SOLVING WITH QUADRATICS

61. 07/12/2018
Example
The distance of an object (d) after a certain amount of time (t), which is accelerating at a rate of ‘A’ metres per second per second (ms-2) and which is already travelling at a speed of u metres per second (ms-1) at time t=0 (i.e. when you start recording) is given by the equation:
d = ut + ½ At2
Calculate how long it takes a car which is already travelling at 80 ms-1 and is accelerating at 8 ms-2 to travel a further 500 metres.
So d = 500, u = 80, A = 8

62. You can’t have –25 seconds!
07/12/2018
Example
So d = 500, u = 80, A = 8
Set up the equation:
From: d = ut + ½ At2
500 = 80t + ½ 8t2
Or: 500 = 80t + 4t2
Rearrange this to give a quadratic equation with 0 on one side of the equals sign
0 = 4t2 + 80t – 500 or 4t2 + 80t – 500 = 0
Factorising gives:
(2t + 50)(2t – 10) = 0
So: 2t + 50 = 0
… or: 2t – 10 = 0
Therefore: 2t = -50
t = -25 seconds
Or: 2t = +10
t = +5 seconds
You can’t have –25 seconds!
This is the answer!

63. 07/12/2018
Example
The unit cost (c) of production of a very exclusive and expensive product is given by the following formula:
c = 75q + 25q2
…where q is the quantity produced.
The company has had a restriction placed (by the Office of Fair Trading) on the selling price of their product. This is going to restrict the cost per unit to £ 250, in order to maintain profits.
So c = 250.
Calculate how many units the firm can produce.

64. 07/12/2018 Example Set up the equation: From: c = 75q + 25q2
Rearrange this to give a quadratic equation with 0 on one side of the equals sign
0 = 25q2 + 75q – or 25q2 + 75q – 250 = 0
Factorising gives:
(5q + 25)(5q - 10) = 0
So: 5q + 25 = 0
… or: 5q - 10 = 0
Therefore: 5q = - 25
q = - 5 units
Or: 5q = +10
q = + 2 units
You can’t have –5 units!
This is the answer!

65. 07/12/2018
Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there.
If her total journey time to work and back is 1 hour, what was her average speed on the way to work?
Remember, time taken =
distance
average speed
Let Jenny’s average speed on the way to work be x.

66. 07/12/2018
Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there.
If her total journey time to work and back is 1 hour, what was her average speed on the way to work?
Jenny’s time taken to get to work = 24 x
Jenny’s time taken to get home from work = 24
x – 20
Total time there and back = 24
x – 20 x +
= 1
Solving this equation will give us the value of x, Jenny’s average speed on the way to work.

67. 07/12/2018
Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there.
If her total journey time to work and back is 1 hour, what was her average speed on the way to work? 24
x – 20 x +
= 1
Start by multiplying through by x(x – 20) to remove the fractions:
24(x – 20) + 24x = x(x – 20)
expand the brackets:
24x – x = x2 – 20x
simplify:
48x – 480 = x2 – 20x
collect terms on the r.h.s.:
0 = x2 – 68x + 480
We have two solutions x = 60 and x = 8.
factorise:
0 = (x – 60)(x – 8)
Which of these solutions is not possible in this situation?

68. 07/12/2018
The only solution that makes sense is x = 60 miles per hour.
If Jenny’s average speed on the way to work was 8 miles per hour her average speed on the way home would be –12 miles per hour, a negative number.
We can therefore ignore the second solution.
When practical problems lead to quadratic equations it is very often the case that only one of the solution will make sense in the context of the original problem.
This is usually because many physical quantities, such as length, can only be positive.

69. 07/12/2018
The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle.
Let’s start by drawing a diagram,
x + 1
x – 7 x
We can use Pythagoras’ Theorem to write an equation in terms of x.

70. 07/12/2018
The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle.
x2 + (x – 7)2 = (x + 1)2
x2 + (x – 7)(x – 7) = (x + 1)(x + 1)
expand:
x2 + x2 – 7x – 7x + 49 = x2 + x + x + 1
simplify:
2x2 – 14x + 49 = x2 + 2x + 1
collect on the l.h.s.:
x2 – 16x + 48 = 0
factorize:
(x – 4)(x – 12) = 0
x = 4 or
x = 12

71. 07/12/2018
The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle.
If x = 4 then the lengths of the three sides are,
4 cm,
4 – 7 = –3 cm
and
4 + 1 = 5 cm
We cannot have a side of negative length and so x = 4 is not a valid solution.
If x = 12 then the lengths of the three sides are,
12 cm,
12 – 7 = 5 cm
and
= 13 cm
So, the shorter sides are 12 cm and 5 cm and the hypotenuse is 13 cm.

72. 07/12/2018
THE DISCRIMINANT

73. Use the quadratic formula to solve 9x2 – 12x + 4 = 0.
07/12/2018
Use the quadratic formula to solve 9x2 – 12x + 4 = 0.
9x2 – 12x + 4 = 0
x =
–b ± b2 – 4ac 2a
x =
2 × 9
12 ±  (–12)2 – (4 × 9 × 4)
x = 18
12 ± 144 – 144
x = 18
12 ± 0
There is only one solution, x = 2 3

74. Use the quadratic formula to solve x2 + x + 3 = 0.
07/12/2018
Use the quadratic formula to solve x2 + x + 3 = 0.
1x2 + 1x + 3 = 0
x =
–b ± b2 – 4ac 2a
x =
2 × 1
–1 ± 12 – (4 × 1 × 3)
x = 2
–1 ± 1 – 12
x = 2
–1 ± –11
We cannot find –11 and so there are no solutions.

75. b2 – 4ac is called the DISCRIMINANT
07/12/2018
The discriminant
From using the quadratic formula,
x =
–b ± b2 – 4ac 2a
we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many solutions there are.
When b2 – 4ac is positive, there are two solutions.
When b2 – 4ac is equal to zero, there is one solution.
When b2 – 4ac is negative, there are no solutions.
b2 – 4ac is called the DISCRIMINANT

76. 07/12/2018
We can demonstrate each of these possibilities using graphs.
Remember, if we plot the graph of y = ax2 + bx + c the solutions to the equation ax2 + bx + c = 0 are given by the points where the graph crosses the x-axis. y x
b2 – 4ac is positive y x
b2 – 4ac is zero y x
b2 – 4ac is negative
Two solutions
One solution
No solutions

77. 12/7/2018