Nathan C. A Level Maths tutor, GCSE Maths tutor, 13 plus  Maths tutor...

Nathan C.

Currently unavailable: until 01/01/2017

Degree: Mathematics (Bachelors) - Warwick University

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About me

About me:

I'm a 3rd Year at Warwick University doing my BSc in Maths. I've tutored since my GCSEs, both online and offline and I absolutely love it! I will make maths enjoyable and understandable, whatever your current level. I promise to also reply to any messages as soon as I get the email too!

Why me?

- Outstanding grades: 8 A*'s, 2 A's for GCSE and 2 A*'s, 2 A's for A level (Maths, Further Maths, Physics and History).

- Deep understanding of Maths: 95.3% average on all maths exams from A level and GCSE, with 100% on 4/6 modules for my maths A level. 

Lots of Experience: Most recently an online tutee of mine went from an E to a B in 3 sessions, right before his AS level exams.

Can make maths fun! (Tutoring siblings tends to make this a necessity!)

- Available at a range of timesI'm either away from University and then free almost every day, or during term time I will have a set schedule which will usually result in evening or weekend tuitions.

The Session:

This is up to you or your child. I can run the session myself, being extremely familiar with the course specification for GCSE and A level, or I can focus on particular topic areas, either those which they struggle at, require more practice, or just want a brief run-over. As the year goes on my sessions will move more to questions and exam papers rather than theory.

My sessions will always be interactive, understandable and friendly. I have tutored so often that I have no problem getting children who are more shy to open up and become confident in their abilities.

Please do not hesitate to get in touch. I look forward to helping you or your child achieve their very best!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr
Maths 13 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA
HistoryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

10/09/2015

Currently unavailable: until

01/01/2017

Questions Nathan has answered

How do I know how many solutions a quadratic equation has?

A quadratic equation is an equation that looks like: x2 + 4x - 2 = 0. The general form of this is written as ax2 + bx + c = 0, where a, b and c are all numbers, and x is our unknown variable. In the example above, we would have a = 1, b = 4 and c = -2. In order to find the number of solution...

A quadratic equation is an equation that looks like:

x2 + 4x - 2 = 0.

The general form of this is written as ax2 + bx + c = 0, where a, b and c are all numbers, and x is our unknown variable. In the example above, we would have a = 1, b = 4 and c = -2.

In order to find the number of solutions, we shall split the quadratic equation into 3 cases.

Case 1: 2 unique solutions - eg x2 + 5x + 6 = 0. Has solutions x = 2 and x = 3.

Case 2: 1 repeated solution - eg x2 + 4x + 4 = 0. Has solution x = 2.

Case 3: No solutions - eg x2 + 2x + 4 = 0. Has no solutions.

But how do we know which case we are in? To do this, we take a look at the quadratic formula, which you will hopefully have seen by now. For reference, it gives the solution of the general quadratic ax2 + bx + c = 0 as:

x = [-b ± √(b2 - 4ac)]/2a

where the ± signifies that the two solutions are 

x = [-b + √(b2 - 4ac)]/2a and 

x = [-b - √(b2 - 4ac)]/2a.

In Case 1, this will give two separate answers for x. In Case 2, both answers will be the same.

However, in Case 3 you will likely arrive at an error! This error arises from the fact that we cannot take the square root of a negative number*. This means, that if we are in case 3, then the section √(b2 - 4ac) is the part that is causing problems! As I said, we cannot take the square root of a negative number, so if b2 - 4ac is negative, we have an error, and no solutions.

This is the key to knowing how many solutions we have: 

If b2 - 4ac is positive (>0) then we have 2 solutions.

If b2 - 4ac is 0 then we have only one solution as the formula is reduced to x = [-b ± 0]/2a. So x = -b/2a, giving only one solution.

Lastly, if b2 - 4ac is less than 0 we have no solutions. 

Example:

How many solutions does x- 3x + 2 = -1 have?

1) Rearrange to fit the general formula: x2 - 3x + 3 = 0. So a = 1, b = -3 and c = 3.

2) Use the formula: b- 4ac = (-3)- 4(1)(3) = 9 - 12 = -3.

3) As b- 4ac < 0, we have no solutions.

So there you have it! Please get in touch if you require any further assistance.

For those interested/advanced students: Technically, you CAN take a square root of a negative number. It's beyond the scope of a GCSE course, so if you're confused by anything after this, don't worry! First of all though, I'll explain why nobody has told you this yet.

Imagine that I asked you to give me the answer to 7 ÷ 3, but you could only use whole numbers. The equation 7 ÷ 3 is equal to 2.33..., but this is not a whole number! So no whole number solutions exist. If I allowed you to use fractions, you could tell me that 7 ÷ 3 is 7/3 or 2 and 1/3.

The same idea applies to the problem here. We only have Real numbers (that is, fractions, decimals, whole numbers and "irrational" numbers such as pi) to deal with the question, and if you are asked to take the square root of a negative number, there are no Real solutions! 

A solution does exist in the "Imaginary" numbers. You don't know about these numbers yet (just like you didn't know about fractions at first). You will learn more about this in A level Further Maths, or perhaps at University, but if this sounds interesting please do check them out via Google.

If b- 4ac < 0 then there are no "Real" solutions. 

However, for your GCSEs, saying that there are no solutions will be good enough for the exam!

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1 year ago

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