Pure Maths How-Tos

(1) You are given an expression stating that a term X is part of this expression

(2) This X is therfore equal to the original expression when multiplied with p(X)

Example

x^3 + 12x^2 + 34 - 12 = (x + 6) x p(x)

• = (x+6) x (Ax^2 + Bx + C)
• = Ax^3 + Bx^2 + Cx + 6x^2 + 6Bx + 6C
• = x^3(A) + x^2(B + 6) + x(6B + C) +6C

Therefore

•  A = 1
• B + 6 = 12 | B = 6
• 6B + C = C + 36 = 34 | C = -2 | (can also be applied to '-12')

P(x) = x^2 + 6x - 2

To do this, you must complete the square on an equation.

Example

{y = x^2 + 4x + 12}   =   {(x + 2)^2 - 2^2 + 12}   =   {(x + 2)^2 + 8}

Vertex

Of (x + 2)^2 + 8, the x value of the vertex is '-1x' what is inside the bracket. They y value is what is outside the bracket

The vertex is (-2, 8)

Line of symmetry

The line of symmetry is sumply what the x equals when on the graph to split the graph in two identical lines. In this case, x = -2

There will be 1, 2 or 0 intersections.

(1) You will be given 2 expressions, which you will need to turn into equations

(2) You will need to find a solution to these simultaneous equations. It may give you co-ordinates: one set, two sets or no sets.

Example

{x + 2y = 3} = {x = 3 - 2y}

{x^2 + 3yx = 10}

Therefore, via substitution, (2y + 1)(y+1) = 0

• When (y= -1/2), ( x= 4)
• When (y= -1), (x= 5)

If the graph's equation >0, you are looking for values above (y= 0)

If the graph's equation <0, you are looking for values below (y= 0)

You are advised to always draw a graph for these questions, as it is very easy to make mistakes.

(1) Factorise to give 2 values of x | use the quadratic formula | use discriminant (for algebraic)

(2) plot these on a graph, highlight the area of the line above or below the x axis, depending on whether the equation >0 or <0

Example

x^2 + 3x - 10 ≥ 0

(x + 5)(x - 2) ≥ 0

x= -5, 2

Equation is above the line and is a positive graph so the inequality is (x≤5), (x≥2)

Some can easily be separated into their constituents and cancelled out.

Others will require more complex methods of division. 2 ways of dividing polynomials are:

• Long division
• The Box Method

Both will give you a quotient, divisor and a remainder

Example

{x^2 + 2x + 1} ÷ {x + 1)

  ____ x   +  1

x+1 |  x^2 + 2x + 1

_____- (x^2 + x)

______________0

Factor Theorem

P(x) = []x^(n) + []x^(n-1) +... []x + []

(1) substitute the divisor in (flip the symbol in the brackets) into p(x)

(2) if it equals 0, it is a factor

(3) find out the rest of the divisors (quotient) to find the rest of the factors for this equation.

Remainder Theorem

P(x) = (x - a)Q(x) + Remainder

This is the same concept as the factor theorm. (x-a) will have its symbol flipped in order for it to go into P(x). This makes the answer {P(a) = 0} and the {remainder = 0}

• Multiplying:  ____________________(x^n)(x^m) = x^(n+m)

• Dividing: ______________________ x^n/x^m =_(x^n-m)

• To the Power of 0: _______________x^0 = 1

• Multiplying by another indice:______(x^n)^m = x^nm

• Indices as a fraction: x^n/m = ______(m√x)^n

• Negative indices: ________________x^-n = 1/x^n

• Base > xª
• _______ ^ index

Exponentials

• graph
• where a > 0
• always passes thrpugh (0, 1)
• steepness increases with greater base
• never touches x axis

Expressions as Logarithms

LogaN = x

a^x = n

e.g Log101000 = 3

10^3 = 1000

• Loga(XY) = LogaX + LogaY

• Loga(X/Y) = LogaX - LogaY

• Loga(X)^n = N(LogaX)

• Loga(1/X) = -LogaX

• LogaA = 1

LogaX = (LogbX) / Logba  

bLogaX = (b) / Logxa

• always use base 10
• state that you are working to base ten (unless specified otherwise)

Taking Logs

take logs using the index, multiplictaion, division and negative laws

e.g 7^(x+1) = 3^(x+2)

(x+1)Log7 = (x+2)Log3

xLog7 + Log7 = xLog3 + 2Log3

x = (2Log2 - Log7) / (Log7 - Log3) [This can be put into a calculator]

{y = ax^3 + bx^2 + cx + d}  =  (x+a)(x+b)(x+c) 

• From here you can find all of the points where the graph crosses the x/y axis
• Can be called a 'repeated root' graph [e.g (-1, 0) (0, 1) (1,0)
• The graph is a 'squiggle' shape
• goes up for positive, and down for negative

Example

y = x^3 - 2x^2 - x + 2

_ = (x+1)(x-1)(x-2)

0 = (-1,0)   (1,0)   (2, 0)

   = 2 [when x=0]

TOTAL CO-ORDINATES: = (-1,0)   (1,0)   (2, 0)  (0, 2)

Plot on a graph

Move in the x axis

| a |     =  [when a>0 = move right]   =  [when a<0 = move left]
| b |     

Move in the y axis

| a |     =   [ when b>0 = move up]   =   [when b<0 = move down]
| b |

Formula

y - b = f(x - a)^n

Example: y = (x-5)^2   by   -4 | 0

y - 0 = (x - 5 - (-4))^2

y = (x-1)^2

n! = n x (n-1) x (n-2)...(n-n+1)

= n factorial

• if you divide a factorial by another factorial, many terms can be cancelled out
• 0! = 1 = (0 - 0 + 1)

N choose R

• n = number of objects
• r = number of different objects that can be chosen from n
• represented by:

(n!) / r!(n - r)! 

Example

| 4 |    =    4! / (3!)(4-3)!  =  24 / 6(1)  = 4
| 3 |

(a + b)^n

• use the relevant row of pascal's traingle (n=x)

Example

The co-efficent of x^2 in the expansion of (2-cx)^3 is 294. Find the value of c.

[a = 2] [b = cx] [n = 3]         look up n = 3

3ab^2

3(2)(cx)^2  =  6(c^2)(x^2)

6c^2 = co-efficient of x^2

6c^2 = 294

c =7

(a + b)^2  =   a^n  +   | n | a^(n-1)b   +   | n | a^(n-2)b^2... +  b^n
___________________| 1 |_____________| 2 |

• in the absence of a calculator, you must do the formula on the previous card ('choosing')

Example

(2x + y)^3

[a= 2x] [b= y] [n= 3]

(2x)^3  +  (3C1)(2x^2)(y)  +  (3C2)(2x)(y^2)  +  (y^3)

=  8x^3  +  6x^2(3y)  +  (12x)(6y^2)  +  (y^3)

• This is to do with terms of a sequence
• terms go:   t1,   t2,   t3...

Example

Un = n(n + 2) - find U3

U3 = (3)(5) = 15

Example

Un-1 = (n-1)(n-1+2) -  find an expression for Un - Un - 1

• imagine the expression without '-1'; it would be (n)(n+2) = Un

= n(n + 2)  - (n - 1)(n-1+2) 

• Inductive sequences are ones that have terms that include the value of the previous term

Un     = Un-1 ...         U1 = U0...

Un+1 = Un+1-1...       U2 = U1...

Example

U1 = 3  |  Un = Un-1  +  2n+1

Find U2

U2 = (3) + (2(2)+1)  =  8

U∞ = L  [this is a lot like taking Logs]

Example

Un+1 = 2 - 1/3Un      [U1 = 3]

Therefore, U2 = 2 - 1/3U1

____________U2 = 2 - 1/3(3) = 1

U∞ = L

L = 2 - 1/3L

4/3L = 2

L = 2/4/3 = 3/2

Un => 1.5   as   n => ∞

• Specific term Un = Un-1 + d  [inductive]
• Nth term = a + (n-1)d

Example

Find the first number after 1000 in the sequence with [a = 6], [d = 2.5]

Un = a + (n-1)d

___= 6 + 5/2(n - 1)

5/2(n - 1) > 994

n - 1 > 397.6

n > 398.6

n = 399, U399 = 1001

Sum of the first n terms

• 1/2 (n)(a + L)

Sum of the first (IN) natural numbers

• 1/2(n)(n+1)

Sum of the sequence

- common difference or first term can be found

- use simultaneous equations 

• (1/2)n [2a + (n-1)d]

_x
∑ n^z
n = y

[x = final value of n]   [y = first value of n]   [z = nth term]

Look at the sequence, find a common term for 'z'. This could be n^x, or dividing each term to make n^x, n^x+1...

Examples

2^2 + 3^2 + 4^2 + 5^2...10^2      [x = 10, y= 2, z= n^2]

100
∑n^2
n = 2

1 + 3 + 9 + 27   =    3^0  +  3^1  +  3^2  +  3^3

x = 27,   y = 1,  nth = n-1

Nth term

• Un = ar^(n-1)

Sum

• Sn = a(1-r^n) / (1 - r)

Example

Find the least number of terms of the geometric series 16 + 20 + 25... required to give a sum greater than 25,000   [a + ar + ar^2...]

20/16 =  5/4

Sn = a(1-r^n)/1-r  = (16(1 - 5/4^n) / 1 - 5/4) > 25000

Cancel out, take logs and solve to find n

n = 27

• a geometric series with common ratio r, converges when | r | < 1
• conergent geometric series has a sum to infinity
  • S∞ = a / (1 - r)

Example

Find the least number of terms of the geometric series 16 + 20 + 25... required to give a sum greater than 25,000

a + ar + ar^2...

20/16 =  

S∞ = a/1-r  = 16 / 1 - 

• The equation may require cancelling down or scaling
• Rearrange the equation so that all terms are equalling 0
• a= the multiplier for x, and b= the multiplier for y 

Example

y = -1/2x - 3

2y = -x - 3

0 = -x - 2y - 3

x + 2y + 3 = 0

• Line Length (Distance beteen 2 points): √(x2- x1) + (y2- y1)

• Midpoint: (x1 +x2)/2, (y1 + y2)/2
  

• Gradient: Change in y / change in x

• Gradient of [OP: a/b] is perpendicular to [OP': -a/b]

• Equation of lime passing through (x1, y1), gradient m: y - y1 = m(x - x1)

Equation of a circle with centre (0, 0), radius r

x^2 + y^2 = r2

The x and y are any set of co-ordinates that lay on the circle

Equation of a circle with centre (a, b), radius r

(x - a)^2 + (y - b)^2 = r^2

For when the circle does have the centre (0, 0) (has 'moved away')

Finding a point outside of a circle circumference

Use the (x-a)^2 + (x-b)^2 = r^2

The outer-circle point is a new 'radius', so just apply its co-ordinates to the formula

Translation = moving a curve without altering its shape by vector | a |
______________________________________________________>.>.| b |

Apply | a | to x^2 + y^2 = r^2
.>.>.>.| b |

This will give you the equation (x - a)^2 + (y - b)^2 = r^2

In general, if a circle is already translated, you just apply the translation further.

(x-p)^2 + (y - q)^2 = r^2 by a translation a b

= (x - p - a)^2 + (y - q - b)^2 = r^2

r^2 does not change

This may involve finding the midpoint (should it be the centre), length of the diameter etc

Example

Circle with traingle P(2, 12), Q(-6, 2) and R(12, -10)

(1) Find midpoint of diameter co-ordinates

(2) Find the distance between midpoint and circumference co-ordinate (radius)

(3) use the equation (x-a)^2 + (y-b)^2 = r^2

• midpoint = [(2+12)/2, (14+-10)/2] = (7, 2)
• distance = √(2-14)^2 + (12--10)^2 = √26
• equation = (x-7)^2 + (x-2)^2 = √26^2

• P = point of tangents meeting
• A = 1 point on the circumference
• B = 1 point on the circumference
• C = centre of the circle

B and A are interchangable

You will be given the co-ordinates 

PA^2 = CP^2 - CA^2

(1) Given the equation of a circle and the co-ordinate of contact

(2) differentiate the equation and apply the co-ordinates to it

(3) this is the gradient

(4) apply the orignal co-ordinates and the new gradient to y-y1 = m(x-x1)

The normal is the same as the tangent, except it is perpendicular, meaning its gradient is -1/m

Example

Curve y=x^2 + 1 on point (3, 10)

Differentiated = 2x     [2(3) = 6]

Tangent equation = y = 6x - 8

Normal equation = y = -1/6x + 10.5

Finding the line of symmetry of a triangle

• Find midpoint of line of equilateral traingle
• find gradient of line between midpoint and point opposite the line
• find equation using gradient and a point on that line

Showing that a triangle has a right angle

• find the gradients of 2 lines
• multiply together
• if it = -1, they are perpendicular, and so must have a right angle between them

Remember: SOH CAH TOA

Sine Rule (length)  =  [a/SinA = b/SinB]  
Sine Rule (angle)  =  [SinA/a = SinB/b] 
Cosine Rule (length)  =  a^2 = b^2 + c^2 - 2bcCosA
Cosine Rule (angle)  =  CosA = b^2 + c^2 - a^2 / 2bc

• Sin30 = 1/2
• Sin45 = 1/(√2)
• Sin60 = (√3)/2
• Cos30 = (√3)/2
• Cos45 = 1/√2
• Cos60 = 1/2
• Tan30 = (√3)/3
• Tan45 = 1
• Tan60 = √3