Currently unavailable: for regular students
Degree: MSci Mathematics (Masters)  University College London University
About Me
Hi!
My name is Pantelis and I am a 4^{th} year student at University College London, in my favorite subject which is Mathematics.
Having already faced the problems of not understanding what the lecturer means, feeling shy to ask for additional info, not knowing who to ask for extra help, I fully appreciate the help and the great difference a good tutor can make to a student’s life.
Tailored to match each student’s personal needs and interests, my tutorials will additionally provide you with the methods and techniques I have personally used, to achieve top grades. The exact same ones I used to complete my Further Mathematics Alevel in just 4 months with Grade ‘A’.
I am generally considered by people who know me as a likeable, calm, patient and flexible person. Hope you will agree too!
The Session:
Depending on your own specific needs, tutorial sessions may include going over the entire material covered at school, concentrating on specific parts you struggle with or looking at and solving exam paper questions.
Maths and Physics are considered difficult subjects and in addition, depending on the Examination Board, exam papers can be a different game in itself.
To assist you in this area, I will demonstrate and make sure you understand:
How past papers are marked
What the raw mark to UMS conversion is
How 23 marks lost due to careless mistakes can lead to a lower grade
Which parts of the material are most likely to come up in the exam
How to use mind maps and flash cards to revise for the exam
How to structure your answers for maximum marks
The Next Step:
The Meet the Tutor Session will be an opportunity for you to let me know what you wish to gain through tutoring, what are you studying and under which examination board, the areas you are facing difficulties with and ultimately if you wish to proceed with tutorials.
If you wish to contact me to arrange a Meet the Tutor Session or for any other enquiry you may have, just fill in the box on the upper right hand side with your enquiry and email address and I will contact you as soon as possible.
CRB/DBS Standard 
No 

CRB/DBS Enhanced 
No 
Please get in touch for more detailed availability
Step 1:
When asked to exrpess something in partial fractions, we first compare the power of the numerator to the power of the denominator.
In our case we have that the power of quadratic equation in the numerator is equal to 2, while the power of the denominator is
(x1)(x2) = x^{2 } 3x +2
which is equal to 2 as well.
Step 2:
Now we devide the numerator by the denominator.
Using long division we get that
(3x^{2}  3x  2)/(x1)(x2) = 3 + (6x8)/(x^{2 } 3x + 2)
Step 3:
The next step is express (6x8)/(x^{2 } 3x + 2) as a partial fraction
(6x8)/(x^{2 } 3x + 2) = (6x8)/(x1)(x2)
(6x8)/(x1)(x2) = A/x1 + B/x2
Step 4:
Now we multiply both the LHS and the RHS by (x1)(x2) because this leads to a common denominator.
6x  8 = A(x2) + B(x1)
Now we have to use two different values for x, such that in the first instance B=0, and in the second instance, A=0
Hence, when x=1,
6(1)  8 = A(12) + B(11)
6  8 = A + 0
2 =  A
A = 2
When x=2
6(2)  8 = A(22) + B(21)
12  8 = 0 + B
B = 4
Step 5:
Now going back to our original equation,
(3x^{2}  3x  2)/(x1)(x2) = 3 + (6x8)/(x^{2 } 3x + 2)
3 + (6x8)/(x^{2 } 3x + 2) = 3 + A/x1 + B/x2
and using A=2 , B=4 we get
3 + (6x8)/(x^{2 } 3x + 2) = 3 + 2/x1 + 4/x2
Hence, our desired result is
(3x^{2}  3x  2)/(x1)(x2) = 3 + 2/x1 + 4/x2
see moreStep 1:
We begin solving this problem by finding the gradient of the tangent line at H, which is equal to dy/dx.
In this case, our curve has equations x = 3sinθ and y= 5cosθ.
Now, we differentiate x and y with respect to θ
dx/dθ = 3cosθ
dy/dθ = 5sinθ
Using the chain rule we get
dy/dx = dy/dθ * dθ/dx
dy/dx = 5sinθ/3cosθ
Now using that at the point H, θ= π/6
dy/dx = 5sin(π/6)/3cos(π/6)
dy/dx = 5(1/2)/3(3^{1/2}/2)
dy/dx = 5/3(3^{1/2})
Step 2:
The gradient of the normal line is equal to  1/M, where M is the gradient of the tangent line at the point H where θ= π/6.
Hence the gradient of the normal line is 3(3^{1/2})/5
Step 3:
Now the general formula for the equation of a line at (X,Y) is y  Y = M(x  X)
In our case we have that, θ= π/6, hence
x=3sin(π/6)= 3/2 and
y=5cos(π/6)= 5(3^{1/2})/2
Hence the equation of the normal line is,
y  5(3^{1/2})/2 = 3(3^{1/2})/5(x  3/2 )
y  5(3^{1/2})/2 =3(3^{1/2})x/5  9(3^{1/2})/10
Multiplying out every term by 10, we are left with
10y  25(3^{1/2}) =6(3^{1/2})x  9(3^{1/2})
10y=6(3^{1/2})x + 16(3^{1/2})
5y=3(3^{1/2})x + 8(3^{1/2})
see moreStep 1:
The fisrt step is to use the following formula when asked to complete the square,
( x + (b/2) )^{2}  (b/2)^{2} + c = 0
Step 2
In our case, b=8,c=24
Hence our equation,
x^{2}  8x + 24= 0
becomes
(x + (8/2) )^{2}  (8/2)^{2} + 24 = 0
which is equal to
(x  4 )^{2}  (4)^{2 }+ 24 = 0
(x  4 )^{2 } 16 + 24 = 0
(x  4 )^{2} + 8 = 0
Step 3:
Now we take 8 to the RHS, as this will allow us to take the square root of both sides.
(x  4 )^{2} = 8
(x  4 ) = +/ (8)^{1/2}
x = 4 +/ (8)^{1/2}
Hence the roots of x^{2}  8x + 24= 0 are
x = 4 + (8)^{1/2} and x = 4  (8)^{1/2}
see more