**D ^{efinition}**

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.

**Q ^{uadratic} I^{nequalities}**;;

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

(-3)^{2}=-3x-3=**+9**

(-6)^{2}=-6x-6=**+36**

If x^{2} is 25, there are two possible answers for x: 5 and -5

This is because (-5)^{2}=25 (negative x negative=positive) and 5^{2}=25

For inequalities containing x^{2} there are positive and negative solutions.

If x is an interger, list all the values which satisfy x^{2}<9

Answer; x= **-3, -2, -1, 0, 1, 2, 3**

If x can have any value solve x^{2}<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;

**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 >)

**Box Plots**- we need to know 5 values to draw a box plot;;

*Lowest Value*

*Lower Quartile/LQ*

*Media*

*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;;

**IQR=UQ-LQ**

So for this example;;

**IQR=38-24
=14**

[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 ^_^

**Q ^{uadratic} S^{equences}**;;

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

**E ^{xample}**;;

1…4…7….10…13…

..3…3….3….3…<-common difference

->n^{th} 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 n^{2} in the n^{th} term for these sequences.

**E ^{xample}**;;

6…13…22…33…46

..7….9….11….13…<- first difference

….2…..2……2…….<-second difference [CONSTANT]

Coefficient of n^{2} in the n^{th} term is 1

Now SUBTRACT n^{2} from each term;;

-n^{2}

1…4…9…16…25

5…9…13…17…21

This sequence has a common difference of 4

The n^{th} term of this sequence is 4n+1

Now put the 2 together

n^{th} term of the sequence;; 6…13…22…33…46

IS

n^^2+4n+1

Ofcourse…CHECK;;

1…2…3…4…5<-n

**O ^{verall} S^{ummary}**;;

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 n^{2} is HALF the second difference

Once we have found the n^{2} bit, SUBTRACT this from the original sequence. We will have a linear left. Find the n^{th} term of this and add it to the n^{2} bit.

**F ^{actorising} Q^{uadratics}**

Factorising means putting brackets back in.

**E ^{xample} (1)**;;

x

^{2}+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!

**E ^{xample} (2)**;;

Factorise x

^{2}-3x-10

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

CHECK!! Expand;;

x^{2}-5x+2x-10=**x ^{2}-3x-10**

**E ^{xample} (3)**;;

Factorise x

^{2}-8x+15

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

CHECK!! Expand;;

x^{2}-5x-3x+15=**x^{2}-8x+15

**S ^{ovling} Q^{uadratics} B^{y} F^{actorising}**;;

Consider the expression x

^{2}+3x-4=0

Factorise it;;

(x+4)(x-1)=0

**GOLDEN RULE**;;

The brackets multiply together to make 0

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

x+4=0…….x=-4

x-1=0…….x=1

**E ^{xample}**;;

x

^{2}-100=0

(x+10)(x-10)=0

**x=-10…OR…x=10**

Now consider THIS equation;;

x^{2}+2x-7=8

"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!

**x ^{2}+2x-15=0** *insert tick*

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

x^{2}+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!

Laterss!