Algebra For The Advanced

Now in this final stage of alegbra from me, Ms A Craig, we will learn how to do Quadratic inequalities&&sequences, Factorising quadratics and solving quadractics by factorising.This is a higher level of algebra and is more difficult than the previous 2 stages, but if you read the definitions and explanations, you shall have a better understanding.

Quadratic Inequalities;;

- When squaring negative numbers, the answer is ALWAYS positive;


If x2 is 25, there are two possible answers for x: 5 and -5
This is because (-5)2=25 (negative x negative=positive) and 52=25

For inequalities containing x2 there are positive and negative solutions.

If x is an interger, list all the values which satisfy x2<9
Answer; x= -3, -2, -1, 0, 1, 2, 3

If x can have any value solve x2<16, then…
x can have any value smaller than 4 or bigger than -4, so x is inbetween -4 and 4
So…ANSWER= -4<x<4

There are a couple of ways we can show this answer, but I am unfortunately unable to illustrate them so I'll just name them instead;

  1. Number Line- a line is drawn on an appropriate scale. You draw a line above the numbers used in the equation. You use an unfiled circle if the number is not inclusive (ie. < or >) and a filled circle if the number is inclusive (ie. < or >)
  1. Box Plots- we need to know 5 values to draw a box plot;;

Lowest Value
Lower Quartile/LQ
Upper Quartile/UQ
Highest Value

If the lowest value on the survey was 2 years and higest was 47years, these will be the end points of the box plot

The interquartile range has 50% of the date within it (ie. the box)

The Inter Quartile Range(IQP) is calculated as follows;;

So for this example;;

[Man this'd be soooo much easier if i could draw the damned thing]

Unfortunately, it's difficult to show ya'll some example questions, but hopefully the above information will help you ^_^

Quadratic Sequences;;

For linear sequences the first difference is always the same. This difference, lets call it a, gives us the coefficient of n in the nth term [check out on //the nth term on a different page, not very algebraic];;

..3…3….3….3…<-common difference

->nth term…..3n+a

The sequence 3n goes like so…3,6,9,12,15…

So a must be -2

What happens when the first difference keeps changing??
ANSWER;; Look at the SECOND difference! DUH!

The second difference (if it is constant) gives us double the coefficient of n2 in the nth term for these sequences.

..7….9….11….13…<- first difference
….2…..2……2…….<-second difference [CONSTANT]

Coefficient of n2 in the nth term is 1

Now SUBTRACT n2 from each term;;

This sequence has a common difference of 4
The nth term of this sequence is 4n+1

Now put the 2 together
nth term of the sequence;; 6…13…22…33…46


Overall Summary;;
When the first difference is always the same, we have a LINEAR SEQUENCE

When the first difference changes, but the second difference is always the same we have a QUADRATIC SEQUENCE.

In a quadratic sequence the coefficient of n2 is HALF the second difference

Once we have found the n2 bit, SUBTRACT this from the original sequence. We will have a linear left. Find the nth term of this and add it to the n2 bit.

Factorising Quadratics
Factorising means putting brackets back in.

Example (1);;
x2+3x+2 factorises to (x+2)(x+1)

This is not easy! We need to look for factors (hence the word "factorise", obviously). We take the art of factorising in the following steps;;

Step One; Look at the constant term (+2). Write out its factor pairs

Step Two; Look for the factor pair which ADD to give the coefficient of x

Step Three; These are the numbers that go into the brackets

NOTE;; This is for the expression where the coefficient of x is 1!

Example (2);;
Factorise x2-3x-10

  1. 1x-10 OR -5x2 OR -1x10 OR -2x5
  2. Factor pair -5 and 2 ADD to give -3
  3. (x-5)(x+2) <-Doesn't matter which way around

CHECK!! Expand;;

Example (3);;
Factorise x2-8x+15

  1. -5x-3 OR -15x-1 OR -3x-5 OR -1x-15
  2. Factor pair -3 and -5 ADD to give -8
  3. (x-5)(x-3)

CHECK!! Expand;;

Sovling Quadratics By Factorising;;
Consider the expression x2+3x-4=0

Factorise it;;

The brackets multiply together to make 0

So ATLEAST one of the brackets must have to be 0, So either;;


Now consider THIS equation;;

"What do I do?!" I hear you cry in panic! Well, you take 8 off both sides so that you ARE left with zero! YAY!

x2+2x-15=0 *insert tick*

When a constant term is 0, you only get ONE bracket;;

x2+4x=0…..TURNS TO…..x(x+4)=0

So either;; x=0…OR…x=-4

Well that's basically it/basically all I can be bothered with in algebra! Hope ya'll enjoyed my extensive hard work which I now have blistered fingers from. GAWD I MUST get a life! Anyways, hope you've enjoyed my teachings!

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