Erin R. GCSE Maths tutor, A Level Maths tutor

Erin R.

Currently unavailable: until 17/01/2016

Degree: Mathematics (Bachelors) - Exeter University

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About me

Hi, I'm Erin, a first year maths student at the University of Exeter. I gained an A* in GCSE and A-Level maths meaning that I have the necessary knowledge to help you reach your full potential.

Maths has always been my favourite subject and I intend to teach you in a way that you will enjoy. In maths the only way to learn and fully understand something is to try the questions yourself, so during my tutorials I will give you plenty of examples that we can work through together. However, I realise that every student is different and so my tutorials will be suited to your individual needs.

The exam board that I studied for both GCSE and A-Level was OCR. However, the maths doesn't tend to differ too much between exam boards so I am available to tutor you even if you are not taking OCR. At A-Level the modules I can help you with are: Core 1, Core 2, Core 3, Core 4, Statistics 1 and Mechanics 1.

When I was studying for my A-Levels I was part of a scheme called Commit to Success that aimed to help GCSE students gain the grade C in maths that they needed in order to study at a higher level or to apply for many jobs. In addition to this I have previously tutored a GCSE student and, coming from a large family, I have had a lot of experience with helping younger siblings with their GCSE and A-Level maths homework. 

If you have any questions then please don't hesitate to contact me through the 'Webmail' facility or book a 'Meet the Tutor' session.

Availability:

During University term time I am available on:
Monday 7-9pm
Tuesday 3-9pm
Wednesday 3-9pm
and the occasional mornings and afternoons on Saturday and Sunday (at request)

During the holidays I will aim to make myself available for any regular students on weekday afternoons/evenings.

I hope I can be of help and look forward to meeting you!

 

 

 

 

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
English Language and LiteratureA-LevelB
ArtA-LevelB
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

17/01/2016

Questions Erin has answered

How do you factorise a quadratic equation?

If you are given a simple quadratic equation, for example x2+6x+8=0, then in order to factorise this you must find two numbers thatadd together to make the coefficient of x in the equation (in this case the coefficient is 6) and multiply together to find the constant (in this case 8). In orde...

If you are given a simple quadratic equation, for example x2+6x+8=0, then in order to factorise this you must find two numbers that add together to make the coefficient of x in the equation (in this case the coefficient is 6) and multiply together to find the constant (in this case 8).

In order to do this you must find the pairs of factors that multiply together to make the constant, so for this example the factors of +8 are 1&8, 2&4, (-1)&(-8) and (-2)&(-4), then using these factors you have to find a pair that will add together to make the coefficient of x, which we know is 6. Therefore the only pair of factors that will add up to 6 are 2 & 4.

So we place these into brackets like so:
(x+2)(x+4)=0

To check your answer you can simply expand the brackets again, which would give you :
x2+2x+4x+8=0
Simplifying to:
x2+6x+8=0, which is what we originally started with therefore showing that we have factorised correctly.

Factorising quadratics can become more complicated when there are negatives or coefficients of xthat are greater than 1. Solving quadratics with negative signs for example: x2-4x-12=0 is done the same way as before. The factors of -12 are: 1&(-12), (-1)&12, 2&(-6), (-2)&6, 3&(-4) and (-3)&4. Then added together the only factors that make -4 are (-6)&2. Therefore the answer would be (x-6)(x+2)=0

They become even more tricky when the coefficient of x2 is greater than 1. For example: 2x2+5x+2=0. You then have to also consider the factors of the coefficient of x2

The only factors of the constant 2 are 1&2 and (-2)&(-1). However one of the factors in a pair will be multiplied by 2, so these factors become: 2&2, 1&4, (-4)&(-1) and (-2)&(-2). We then need to find out which of these factors add up to 5, which we can see is 1&4.

So our final answer is (2x+1)(x+4).

 

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2 years ago

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