# Maths - Half Term 1 - Examples and Formulae

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Scale Drawings:

• A scale drawing = an accurate representation of a real object.
• To find the measurements for a scale drawing: Multiply all the actual measurements by a common scale factor.
• Scale factors are often ratios e.g. 1cm:1m
• If there are no units they are the same e.g. 1cm:100cm = 1:100
• Map scales could by 1:50,000 or 1:200,000
• This means 1cm equates to 50,000cm or 500m in real life and 200,000cm or 2km in real life

x2+6x+5=0
1. Find 2 numbers which add to the middle number (b) and multiply to the end number (c).
2. x2+6x+5=0 is the formula ax2+bx+c
3. In this case the 2 numbers would be 5 and 1
4. Then we re-write the equation as (x+5)(x+1)=0
5. Now we can split it into (x+5)=0 and (x+1)=0
6. Therefore x=-5 and x=-1

Difference of two squares

Ex1.                                 x2-25=0

square - negative - square

• Can be factorised to form 2 factors

x2 = 25

(x+5)(x-5)=0

x=-5 or x=5

check: (x+5)(x-5)=0

x2-5x+5x-25

x2-25 = correct as -5x+5x cancels out

Ex2.                                  9x2-16=0

9x2 = 16

(3x+4)(3x-4)=0

3x+4=0                    3x-4=0

3x=-4.                     3x=4

x=-4/3                    x=4/3

Solving Quadratic Equations by Factorisation of ax2+bx+c where a doesn't equal 1

ax2+bx+c=0

E.g. 2x2-3x-2=0

1. Multiply 'a' and 'c': 2x-2=-4

2. Find 2 numbers that multiply to 'ac' and add to 'b': -4 and 1

3. Rewrite equation with new numbers: 2x2-4x+1x-2=0

4. Split down the middle: 2x2-4x     1x-2=0

5. Factorise each side: 2x(x-2) + 1(x-2) = 0

6. Collect 'like terms' in the brackets: (2x+1)(x-2)=0

7. Solve like normal: 2x+1=0   2x=-1   x=-1/2

x-2=0   x=2

Completing the Square

• 2 uses:
• 1. Finding a minimum or maximum of a quadratic curve
• 2. To find a solution to a quadratic equation

E.g.                             x2+6x-16=0

• We use completing the square because factorisation won't work

1. Focus on the first 2 terms: x2 - 6x

2. Replace with a squared bracket of sqrt first value and half second value: (x-3)2

3. Subtract the square of the halfed second value: (x-3)2-9

4. Reintroduce the other terms: (x-3)2-9-16=0

5. Simplify: (x-3)2-25=0

6. Solve: (x-3)2=25

x-3= 5 or