Gemma L. GCSE Maths tutor, A Level Maths tutor, GCSE Physics tutor, A...

Gemma L.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Bath University

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About me

Hi everyone, I'm Gemma. At the moment I am studying (hard) for a Mathematics degree at Bath University. I heave always had a deep love for all things Maths related since I started learning the subject at school. I hope to be able to share my knowledge, experience and passion with you! I also have a keen interest in Physics and did well at school and A level, so if you are in need of some help, I'm your girl!

Through our tutorial sessions, we can build your confidence in Maths/Physics whilst you gain the skills and understanding desired by teachers and examiners to ace any question they give you.

Before

If you are interested in my tutoring, drop me a message to see when I am free to do a Meet-the-Tutor session. If you could include in your message what subject you would like tutoring for, and at what level. Also, if possible, how far along in your studies you are and anything in particular you are struggling with. (Or just keep it brief lol)

We can then organise a session to get to know each other and use this time to foresee how our first tutoring session will go.

During

It would be most beneficial if we decided at the end of each tutorial what you wanted to go over in the next. This will give me time to prepare examples, questions, visuals etc. to give you quality learning assistance. I will therefore welcome you to come to the tutorials with questions/queries related to the topics we have decided on. 

We can go over exam questions if you would like, but I would like to focus on your understanding of concepts so you'll be able to answer questions on your own.

Finally

I look forward to hearing from you! Message me any time and I will get back to you as quick as I can.

I'm flexible with when I can tutor.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Please get in touch for more detailed availability

Questions Gemma has answered

How do I factorise a quadratic equation?

Firstly, make sure you agree that a quadratic equation is an equation of the form y = ax2 + bx + c where a,b,c are (real) constants (a is not 0), and x is the variable. (Note: the equation has an "x squared" term). An example would be: y = x2 + x - 6. We want to factorise this equation so we ...

Firstly, make sure you agree that a quadratic equation is an equation of the form y = ax2 + bx + c where a,b,c are (real) constants (a is not 0), and x is the variable. (Note: the equation has an "x squared" term).

An example would be: y = x2 + x - 6.

We want to factorise this equation so we can find the values of x for which y = 0 (the points where the curve crosses the x-axis).

Secondly, we need to be aware that factorising a quadratic mens expressing the equations as a product of two brackets which each contain an term.

So, in the form ax2 + bx + c = (dx + e)(fx + g)

Working with our example, y = x2 + x - 6, we first direct our attention to the constant term c, in this case c = -6.

If we expand the brackets, we get ax2 + bx + c =(dx+e)(fx + g) = dfx2 + (ef +dg)x + eg.

We should already be able to see that for our example where the coefficient of x2 is 1 that d,f = 1 so now we have a simpler equation: 

(x+e)(x+g) = x2 + (e + g)x + eg. (1.1)

We can then use eg = c = -6, our constant term ie. the constant terms in the brackets multiple to make our origanal constant term.

To find e,h, we think of pairs of numbers which multiply to give -6 :

-1 x 6 // -2 x 3 // -3 x 2 // -6 x 1

So how do we decide which pair of number will give the correct equation?

Well we could test each pair and multiply out the brackets until we get the right equation, but this could take some time if we have more than four options, so instead we'll take a shortcut:

See which pairs ​add to give the coefficient of x

From (1.1), we can equate e + g = b.

-1 + 6 = 5 // -2 + 3 = 1 // -3 + 2 = -1 // -6 + 1 = -5

In our example, b = 1, so we can tell that our constants, e,h are -2,+3 and our answer is: y = (x - 2)(x + 3).

Check! (x - 2)(x + 3) = x2 - 2x + 3x - 6 = x2 + x - 6 

as required!

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8 months ago

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