1. Applications of Quadratic Equations

2. GEOMETRY
The top of a coffee table is 3 feet longer than it is wide and has an area of 10 square feet. What are the dimensions of the top of the coffee table?
5 feet
L = w + 3
Let's draw a picture:
Let width = w w
2 feet
Area = length x width
10 = (w+3)w
Solve this by multiplying out and getting everything on one side = 0 and factoring
0 = w2 + 3w - 10
w = -5 or w = 2
L = = 5 feet
0 = (w +5)(w-2)
Since width can't be negative throw out –5 and width is 2 feet
w + 5 = 0 or w – 2 = 0

4. 500 500 COMPOUND INTEREST P = $500 and A = $572.45
Amount in account after two years
Interest rate as a decimal
Principal Amount you deposit
P = $500 and A = $572.45
Let's substitute the values we are given for P and A
Solve this equation for r
500
500
Square root both sides but don't need negative because interest rate won't be negative

5. PYTHAGOREAN THEOREM
An L-shaped sidewalk from building A to building B on a college campus is 200 feet long. By cutting diagonally across the grass, students shorten the walking distance to 150 feet. What are the lengths of the two legs of the sidewalk?
200-x
Draw a picture: B x
If first part of sidewalk is x and total is 200 then second part is x
150 A
Using the theorem:
Multiply out
continued on next slide

6. use the quadratic formula to solve
get everything on one side = 0
divide all terms by 2
use the quadratic formula to solve 1
= 64.6 so doesn't matter which you choose, the two lengths are meters and 64.6 meters.

7. = + + = WORK-RATE PROBLEM
An office contains two copy machines. Machine B is known to take 12 minutes longer than Machine A to copy the company's monthly report. Using both machines together, it takes 8 minutes to reproduce the report. How long would it take each machine alone to reproduce the report?
Work done by Machine A
Work done by Machine B =
1 complete job +
Rate for A
1 over time to complete alone
Rate for B
1 over time to complete alone
Time to complete job
Time to complete job + = 1
Call t time for machine A to complete
(continued on next slide)

8. Clear the equation of fractions by multiplying all terms on both sides by the common denominator and cancel all fractions.
Get everything on one side = 0 and factor
Throw out –8 because negatives don't make sense as a time to complete the job
So Machine A can complete the job alone in 12 minutes and Machine B would take or 24 minutes.

9. After how many seconds will the height be 11 feet?
Height of a tennis ball
A tennis ball is tossed vertically upward from a height of 5 feet according to the height equation
where h is the height of the tennis ball in feet and t is the time in seconds.
After how many seconds will the height be 11 feet?
So there are two answers: (use a calculator to find them making sure to put parenthesis around the numerator) t = .42 seconds or .89 seconds.
Get everything on one side = 0 and factor or quadratic formula.
-11
-11

10. When will the tennis ball hit the ground?
What will the height be when it is on the ground? h = 0
So there are two answers: (use a calculator to find them)
t = or 1.52 seconds (throw out the negative one)

11. Average Speed
A truck traveled the first 100 miles of a trip at one speed and the last 135 miles at an average speed of miles per hour less. If the entire trip took 5 hours, what was the average speed for the first part of the trip?
If you used t hours for the first part of the trip, then the total 5 minus the t would be the time left for the second part.
Let's make a table with the information
first part
second part
distance
rate
time r
100 t
135
r - 5
5 - t

12. Use this formula to get an equation for each part of trip
Distance = rate x time
distance
rate
time r
first part
100 t
second part
135
r - 5
5 - t
Solve first equation for t and substitute in second equation
100 = r t
135 = (r - 5)(5 - t) r r

13. r r r r r FOIL the right hand side
Multiply all terms by r to get rid of fractions r r r r
Combine like terms and get everything on one side
Divide everything by 5
Factor or quadratic formula
So r = 50 mph since r = 2 wouldn't work for second part where rate is r –5 and that would be –3 if r was 2.

14. The following series of lessons will equip you with the necessary skills to complete a complex investigation at the end of the unit.
The initial lessons may seem tedious, but bear with us…

15. At the end of the unit you will be presented with this problem:x
What size must the cut-out corners be to give the maximum volume for the open box?

16. To attempt this problem you need to be able to:
LESSON 1
Simplify algebraic expressions x
LESSON 2
Solve algebraic equations
LESSON 3
Formulate your own
expressions & equations
Solve equations by
trial & improvement
LESSON 4
Accurately rearrange formulae
LESSON 5

17. Objective: Monday 21st February
To be able to simplify algebraic expressions
Level 5/6
Lesson 1

18. About 500 meteorites strike Earth each year.
A meteorite is equally likely to hit anywhere on earth.
The probability that a meteorite lands in the
Torrid Zone is
Area of Torrid Zone
Total surface area of earth
Let r represent the earth’s radius. Write an expression to estimate the area of the torrid zone.
You can think of the distance b/w the tropics as the height of a cylindrical belt around the earth at the equator.
The length of the belt is the earth’s circumference 2пr.
The surface area of a sphere is 4пr2.Write and simplify expressions for the area of,and probability that a meteor lands in, the torrid zone.
Use 3963 miles for the earth’s radius. Find the probability.

19. Take the number of the month of your birthday…
Multiply it by 5
Add 7
Multiply by 4
Add 13
Multiply by 5
Add the day of your birth
Subtract 205

20. What have you got? Why does this work? Homework:
Write an algebraic expression and simplify it to prove why this works.

21. The length of a rectangular field is a metres.
The width is 15m shorter than the length.
The length is 3 times the width. a
a - 15
Write down an equation in a and solve it to find the length and width of the field.
Length = 22.5m Width = 7.5m

22. Imagine a triangle
Choose a length for its base.
Call it ‘z’
Make the vertical height 3 units longer than the base
Work out the area of your triangle
Write down an equation in z that satisfies your conditions
Give it to your partner to solve for the base length of your triangle (z).

23. Problems involving quadratic equations
A rectangle has a length of ( x + 4) centimetres and a width of ( 2x – 7) centimetres.
If the perimeter is 36cm, what is the value of x?
X = 7
x + 4
2x - 7
If the area of a similar rectangle is 63cm2 show that 2x2 + x – 91 = 0 and calculate the value of x
X = 6.5

24. Monday 28th February
Objective:
Formulate equations and solve by trial and improvement.
Level 6 / 7

25. The length of a rectangular field is a metres.
The width is 15m shorter than the length.
The length is 3 times the width. a
a - 15
Write down an equation in a and solve it to find the length and width of the field.
a = 3 (a – 15)
Length = 22.5m Width = 7.5m

26. Formulating quadratic equations
Joan is x years old and her mother is 25 years older.
The product of their ages is 306.
a) Write down a quadratic equation in x
b) Solve the equation to find Joan's age.
x ( x + 25 ) = 306
x2 + 25x = 306
x x – 306 = 0
How can we solve this?
Factorisation?
Graphically
Formula

27. A more accurate method is
x x – 306 = 0
This example factorises:
( x + 34 )( x – 9 ) = 0
Either x + 34 = 0
Or, x – 9 = 0
x = -34
x = 9
Since Joan cannot be –34 years old, she must be 9.
Some quadratic equations do not factorise exactly.
Solving some equations (i.e. cubic ) by a graphical method is not very accurate.
A more accurate method is
trial and improvement

28. Solving equations by trial and improvement.
E.g. 1
A triangle has vertical height 3 cm longer than its base.
It’s area is 41 cm2. What is the length of its base to 1 d.p? 41
x ( x + 3) = 41 2
x +3
x ( x + 3) = 41 x 2 = 82
x2 + 3x – 82 = 0 x
Too small
Try x = (3 x 7) – 82 = -12
Too big
Try x = (3 x 8) – 82 = 6
Too small
Try x = ( 3 x 7.6) – 82 =
Try x = ( 3 x 7.7) – 82 = 0.39
Too big
Too small
Try x = ( 3 x 7.65) – 82 =
x = 7.7 to 1 d.p
Base = 7.7cm to 1 d.p

29. 5x2 – 12x + 5 = 0 For x > 1 x2 – 5x – 1 = 0 For x > 0
Race groups with different methods of solving I.e. formula, trial & imp, factorising…
1.9
5.2
1.8
0.5
To 1 d.p

30. Transposition of formula
Wednesday 2nd March
Transposition of formula
Objective:
To be able to accurately rearrange formula for a given subject.

31. Here are some questions and answers (by students A and B) on rearranging formulae.
Decide which answers to tick (correct) and which to trash (incorrect).
You must give reasons for your decision.
Question 2. Daniel buys n books at £4 each. He pays for them with a £20 note. He receives C pounds in change.
Write down a formula for C in terms of n.
Question 1.
Make x the subject of the following:
Y = x2 + 4 5
Student A answer
Y = x2 + 4 5
5y = x2 + 4
5y – 4 = x2
x2 = 5y – 4
x = 5y - 4 x
Student B answer
Y = x2 + 4 5
5y = x2 + 4
x2 + 4 = 5y
x2 = 5y – 4
x =
5y – 4
Trash
Trash
Books cost £4n
Change C = £4n - 20
Change = £20 – cost of books
C = 20 – 4n

32. Rearrangement of formulae
When doing these sort of problems, remember these things:
a) Whatever you do to one side of the formula, you must also do the same to the other side:
To rearrange the following formula making x the subject
Add y to both sides of the formula giving:
As (–y + y = 0) and (2y + y = 3y) we can say:
Now subtract x from both sides leaving:
So to get x we can now divide both sides by 2:

33. b) When you are dealing with more complicated formulae, try to strip off the outer layers first.
First get rid of the square root, by squaring both sides
Now get rid of the division bar, by multiplying both sides by x
To leave you with x on one side, divide both sides by g2
c) When you want to get rid of something in a formula, remember to do the opposite (inverse) to it.

34. One last example…
Make u the subject of the following:
Multiply both sides by (u + v)
Expand the bracket
Collect the u terms on one side
Factorise the LHS to isolate u
Divide both sides by (f – v)

35. Applications of Quadratic Equations