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.

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.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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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

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

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