Using and applying Using and applying mathematicsmathematics

Sequences & FormulaeSequences & Formulae

Year 10Year 10

The following series of lessons will The following series of lessons will equip you with the necessary skills equip you with the necessary skills to complete a complex investigation to complete a complex investigation at the end of the unit.at the end of the unit.

The initial lessons may seem The initial lessons may seem tedious, but bear with us…tedious, but bear with us…

At the end of the unit you will be At the end of the unit you will be presented with this problem:presented with this problem:

x

x

What size must the cut-out corners be to give the maximum volume for the open box?

LESSON 2

LESSON 3

LESSON 1

LESSON 4

LESSON 5

To attempt this problem you need to be able to:

x

x Simplify algebraic expressions

Solve algebraic equations

Formulate your own expressions & equations

Solve equations by trial & improvement

Accurately rearrange formulae

Objective:Objective:

To be able to simplify algebraic To be able to simplify algebraic expressionsexpressions

Level 5/6

Monday 21st February

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

Objective:Objective:

To be able to solve algebraic To be able to solve algebraic equations. equations.

Level 6/7

Wednesday 23rd February

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

What have you got?What have you got?

Why does this work?

Homework:

Write an algebraic expression and simplify it to prove why this works.

Objective:Objective:

To be able to formulate expressions To be able to formulate expressions and formulae.and formulae.

Thursday 24th February

Level 6/7

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

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).

Problems involving quadratic equationsProblems 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 + 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 = 7

X = 6.5

Monday 28th February

Objective:

Formulate equations and solve by trial and improvement.

Level 6 / 7

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

a = 3 (a – 15)

Formulating quadratic equationsFormulating quadratic equationsJoan 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

x2 + 25x – 306 = 0

How can we solve this?

Factorisation?

Formula

Graphically

x2 + 25x – 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

Solving equations by trial and improvement.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

x ( x + 3) = 41

2

x ( x + 3) = 41 x 2 = 82

x2 + 3x – 82 = 0

Try x = 7 72 + (3 x 7) – 82 = -12

Too small

Try x = 8 82 + (3 x 8) – 82 = 6 Too big

Try x = 7.6 7.62 + ( 3 x 7.6) – 82 = - 1.44

Too small

Try x = 7.7 7.72 + ( 3 x 7.7) – 82 = 0.39 Too big

Try x = 7.65 7.652 + ( 3 x 7.65) – 82 = -0.5275Too small

x = 7.7 to 1 d.p

Base = 7.7cm to 1 d.p

a) 5x2 – 12x + 5 = 0 For x > 1

b) x2 – 5x – 1 = 0For x > 0

c) 2x2 – 2x – 3 = 0 For x > 0

d) 5x2 + 9x – 6 = 0 For x > 1

To 1 d.pa) 1.9

b) 5.2

c) 1.8

d) 0.5

Transposition of formula

Objective:

To be able to accurately rearrange formula for a given subject.

Wednesday 2nd March

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 1.

Make x the subject of the following: Y = x2 + 45

Student A answer

Y = x2 + 45

5y = x2 + 4

5y – 4 = x2

x2 = 5y – 4

x = 5y - 4x

Student B answer

Y = x2 + 45

5y = x2 + 4

x2 + 4 = 5y

x2 = 5y – 4

x = 5y – 4

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.

Books cost £4n

Change C = £4n - 20

Change = £20 – cost of books

C = 20 – 4n

Rearrangement of formulaeRearrangement of formulaeWhen 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:

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.

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)