Bogdan-Adrian M. A Level Maths tutor

Currently unavailable: for regular students

Degree: Mathematics and Computer Science (Masters) - Bristol University

MyTutor guarantee

About me

I am an undergraduate, Mathematics and Computer Science student at the University of Bristol. In 2014 I graduated from "Petru Rares" National College passing my Baccalaureate exam with 10/10 in Mathematics. I participated in numerous mathematics competitions and Olympiads; my best result was achieved in the 2014 "National Olympiad in Mathematics" where I obtained a Bronze Medal. In April 2013 I successfully attended a program called "Teacher for a day" on mathematics.

My expertise areas include topics such as Number Teory and Group Theory, Calculus & Analysis, Algebra, Complex Numbers, Linear Algebra (in particular vectors and matrices) and Combinatorics (specially elements of Graph Theory). As far as the tutorials are concerned I am happy to explain key concepts and useful results, solve problems of various levels of difficulty, check the solutions proposed by the student and compare them with model solutions, or anything else required.

Mathematics, apart from being extremely useful in our daily life, it is a very beautiful subject. Part of my task will be to help you see its beauty behind coplicated equations and, so called, "hard to remember" formulae. Once that is accomplished, problems will become exercises and exercises will become an easy and enjoyable task.

Subjects offered

Subject	Qualification	Prices
Maths	A Level	£20 /hr
Maths	GCSE	£18 /hr

Qualifications

Subject	Qualification	Grade
Mathematics	International Baccalaureate (IB)	10
Programming	International Baccalaureate (IB)	9.95
Romanian Language and Literature	International Baccalaureate (IB)	9.60

CRB/DBS Standard		No
CRB/DBS Enhanced		No

General Availability

Currently unavailable: for regular students

Weeks availability

	Mon	Tue	Wed	Thu	Fri	Sat	Sun

Weeks availability

	Before 12pm	12pm - 5pm	After 5pm
MONDAY
TUESDAY
WEDNESDAY
THURSDAY
FRIDAY
SATURDAY
SUNDAY

Please get in touch for more detailed availability

Questions Bogdan-Adrian has answered

Given that (cos(x)^2 + 4 sin(x)^2)/(1-sin(x)^2) = 7, show that tan(x)^2 = 3/2

First, we use 1 - sin(x)^2 = cos(x)^2 and get: (LHS) (cos(x)^2 + 4 sin(x)^2)/(1-sin(x)^2) = (cos(x)^2 + 4 sin(x)^2)/cos(x)^2 = 1 + 4 (sin(x)/cos(x))^2 = 1 + 4 tan(x)^2 Now we know that the left hand side is equal to 7. Hence, 1 + 4 tan(x)^2 = 7 <=> tan(x)^2 = 3/2

First, we use 1 - sin(x)^2 = cos(x)^2 and get:

(LHS) (cos(x)^2 + 4 sin(x)^2)/(1-sin(x)^2)

= (cos(x)^2 + 4 sin(x)^2)/cos(x)^2

= 1 + 4 (sin(x)/cos(x))^2

= 1 + 4 tan(x)^2

Now we know that the left hand side is equal to 7.

Hence, 1 + 4 tan(x)^2 = 7 <=> tan(x)^2 = 3/2

2 years ago

607 views

If y = 1/(x^2) + 4x, find dy/dx

First, notice that 1/(x^2) = x^(-2) dy/dx = d/dx (x^(-2) + 4x) The derivative of the sum is the sum of the derivatives = d/dx (x^(-2)) + d/dx (4x) The derivative of x^n is nx^(n-1), for every real number n, and a constant gets in front of the deivative = -2 x^(-3) + 4 d/dx (x) = -2/(x^3) + 4

First, notice that 1/(x^2) = x^(-2)

dy/dx = d/dx (x^(-2) + 4x)

The derivative of the sum is the sum of the derivatives

= d/dx (x^(-2)) + d/dx (4x)

The derivative of x^n is nx^(n-1), for every real number n, and a constant gets in front of the deivative

= -2 x^(-3) + 4 d/dx (x)

= -2/(x^3) + 4

2 years ago

604 views

The quadratic equation 2x^2 + 8x + 1 = 0 has roots x1 and x2. Write down the value of x1+x2 and x1*x2 and find the value of x1^2 + x2^2

If a quadratic equation is of the form : ax^2 +bx +c =0 and has roots x1 and x2, then the following statements are true: x1+x2 = - b/a x1*x2 = c/a In our case: x1+x2 = - 8/2 = -4 and x1*x2 = 1/2 Now, x1^2 + x2^2 = (x1+x2)^2 -2*x1*x2 = (-4)^2 -2 * 1/2 = 16 - 1 = 15

If a quadratic equation is of the form : ax^2 +bx +c =0

and has roots x1 and x2, then the following statements are true:

x1+x2 = - b/a

x1*x2 = c/a

In our case: x1+x2 = - 8/2 = -4

and x1*x2 = 1/2

Now, x1^2 + x2^2 = (x1+x2)^2 -2*x1*x2

= (-4)^2 -2 * 1/2 = 16 - 1 = 15

2 years ago

615 views

Send a message