Hi there! I’m **George**. I'm a **University of Warwick graduate** with a **first-class Master of Mathematics degree**. I tutor **GCSE** and **A-Level** students who are **dedicated to learning** and **willing to put in the effort to succeed**!

🌟 **What differentiates you from other tutors?** 🌟

📙 I actually tutor for a living! For over three years I've been tutoring on MyTutor with fantastic results and feedback from both students and parents

📘 I solely tutor Maths, meaning I have immense experience in tutoring for various qualifications and exam boards for the subject

📗 Whilst at Warwick I was a supervisor to two groups of undergraduate Maths students; I mentored them throughout their first year and had a blast doing so!

🌟 **Why did you study Maths? **🌟

I'm a problem solver. When I approach a puzzle I strive to master it. If things get tough, I don't just throw in the towel. Instead, I try approaching the problem from different perspectives, applying different methods, or maybe even tackling special cases, just to gain that little bit of insight. And at the end of this slow and sometimes painful process comes immense satisfaction knowing that you've conquered the problem.

Hi there! I’m **George**. I'm a **University of Warwick graduate** with a **first-class Master of Mathematics degree**. I tutor **GCSE** and **A-Level** students who are **dedicated to learning** and **willing to put in the effort to succeed**!

🌟 **What differentiates you from other tutors?** 🌟

📙 I actually tutor for a living! For over three years I've been tutoring on MyTutor with fantastic results and feedback from both students and parents

📘 I solely tutor Maths, meaning I have immense experience in tutoring for various qualifications and exam boards for the subject

📗 Whilst at Warwick I was a supervisor to two groups of undergraduate Maths students; I mentored them throughout their first year and had a blast doing so!

🌟 **Why did you study Maths? **🌟

I'm a problem solver. When I approach a puzzle I strive to master it. If things get tough, I don't just throw in the towel. Instead, I try approaching the problem from different perspectives, applying different methods, or maybe even tackling special cases, just to gain that little bit of insight. And at the end of this slow and sometimes painful process comes immense satisfaction knowing that you've conquered the problem.

In my tutorials the** student is in charge**: you **control** what we cover.

🌟 **During tutorials we can: **🌟

📙 Go through **sections of your course** that you're struggling with

📘 Look at **particular questions** that you can’t get your head around

📗 **Identify areas of difficulty** by going through past paper questions and build on them

🌟 **Structure of Tutorials **🌟

📙 I will **introduce** and **explain** new **definitions and concepts**

📘 We will work through several examples **together** and you'll have the opportunity to practice similar questions **by yourself** to help **build your confidence**

📗 We will then **consolidate the principles** that we have learned by tackling **past paper questions**

**Refund policy:** Tutorials can be cancelled and refunded any time up to 24 hours before the tutorial is due to start.

**Coursework:** I don't offer help on coursework that will contribute to your final grade.

In my tutorials the** student is in charge**: you **control** what we cover.

🌟 **During tutorials we can: **🌟

📙 Go through **sections of your course** that you're struggling with

📘 Look at **particular questions** that you can’t get your head around

📗 **Identify areas of difficulty** by going through past paper questions and build on them

🌟 **Structure of Tutorials **🌟

📙 I will **introduce** and **explain** new **definitions and concepts**

📘 We will work through several examples **together** and you'll have the opportunity to practice similar questions **by yourself** to help **build your confidence**

📗 We will then **consolidate the principles** that we have learned by tackling **past paper questions**

**Refund policy:** Tutorials can be cancelled and refunded any time up to 24 hours before the tutorial is due to start.

**Coursework:** I don't offer help on coursework that will contribute to your final grade.

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.

Enhanced DBS Check

09/01/2018This refers to a question in the Edexcel GCSE paper this year which took students to Twitter venting their frustration. The question is as follows:

⠀

**‘There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.Hannah takes a random sweet from the bag. She eats the sweet.Hannah then takes at random another sweet from the bag. She eats the sweet.The probability that Hannah eats two orange sweets is 1/3.Show that n² – n – 90 = 0’**

⠀

Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer.

⠀

Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line. So let’s do that.

⠀

What’s the probability that the first sweet she eats from the bag is orange? There are**n** sweets in the bag, **6** of which are orange. So the probability is 6/n. ⠀

⠀

What is the probability that the second sweet she eats from the bag is orange? Now there are**n-1** sweets in the bag, **5** of which are orange (since she has eaten an orange sweet!). So the probability is 5/(n-1).⠀

⠀

These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n-1) = 30/n(n-1).⠀

⠀

But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.⠀

⠀

30/n(n-1) = 1/3

⠀

⇒ 90/n(n-1) = 1

⠀

⇒ 90 = n(n-1)

⠀

⇒ n(n-1) – 90 = 0

⠀

⇒ n² – n – 90 = 0 ⠀

⠀

Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!This refers to a question in the Edexcel GCSE paper this year which took students to Twitter venting their frustration. The question is as follows:

⠀

**‘There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.Hannah takes a random sweet from the bag. She eats the sweet.Hannah then takes at random another sweet from the bag. She eats the sweet.The probability that Hannah eats two orange sweets is 1/3.Show that n² – n – 90 = 0’**

⠀

Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer.

⠀

Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line. So let’s do that.

⠀

What’s the probability that the first sweet she eats from the bag is orange? There are**n** sweets in the bag, **6** of which are orange. So the probability is 6/n. ⠀

⠀

What is the probability that the second sweet she eats from the bag is orange? Now there are**n-1** sweets in the bag, **5** of which are orange (since she has eaten an orange sweet!). So the probability is 5/(n-1).⠀

⠀

These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n-1) = 30/n(n-1).⠀

⠀

But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.⠀

⠀

30/n(n-1) = 1/3

⠀

⇒ 90/n(n-1) = 1

⠀

⇒ 90 = n(n-1)

⠀

⇒ n(n-1) – 90 = 0

⠀

⇒ n² – n – 90 = 0 ⠀

⠀

Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!

⠀

⠀

Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer.

⠀

Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line. So let’s do that.

⠀

What’s the probability that the first sweet she eats from the bag is orange? There are

⠀

What is the probability that the second sweet she eats from the bag is orange? Now there are

⠀

These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n-1) = 30/n(n-1).⠀

⠀

But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.⠀

⠀

30/n(n-1) = 1/3

⠀

⇒ 90/n(n-1) = 1

⠀

⇒ 90 = n(n-1)

⠀

⇒ n(n-1) – 90 = 0

⠀

⇒ n² – n – 90 = 0 ⠀

⠀

Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!This refers to a question in the Edexcel GCSE paper this year which took students to Twitter venting their frustration. The question is as follows:

⠀

⠀

Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer.

⠀

Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line. So let’s do that.

⠀

What’s the probability that the first sweet she eats from the bag is orange? There are

⠀

What is the probability that the second sweet she eats from the bag is orange? Now there are

⠀

These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n-1) = 30/n(n-1).⠀

⠀

But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.⠀

⠀

30/n(n-1) = 1/3

⠀

⇒ 90/n(n-1) = 1

⠀

⇒ 90 = n(n-1)

⠀

⇒ n(n-1) – 90 = 0

⠀

⇒ n² – n – 90 = 0 ⠀

⠀

Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!

This is a question from an FP1 paper. Here i denotes √-1.

⠀

Fact that should be burned into your soul:**‘Complex roots of a polynomial equation with real coefficients form conjugate pairs’**.

⠀

This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and -1 – 2i are also roots of P(z) = 0. So we now have 4 roots of our**quartic **equation P(z) = 0, so that’s all of them!

⠀

We can now employ the**factor theorem**. Remember, this states that if z = α is a root of a polynomial then (z – α) is a factor of that polynomial.

⠀

So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that P(z) = (z – (2 - i))(z – (2 + i))(z – (-1 + 2i))(z – (-1 - 2i))

⠀

So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.

⠀

⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)

⠀

⇒ P(z) = (z² - 2z – zi – 2z + 4 + 2i + zi - 2i - i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i - 4i²)

⠀

⇒ P(z) = (z² - 4z + 5)(z² + 2z + 5) [Remember: i² = - 1]

⠀

⇒ P(z) = z⁴ + 2z³ + 5z² - 4z³ - 8z² - 20z + 5z² + 10z + 25

⠀

⇒ P(z) = z⁴ - 2z³ + 2z² - 10z + 25

⠀

Thus a = -2, b = 2, c = -10, d = 25 and we’re done.This is a question from an FP1 paper. Here i denotes √-1.

⠀

Fact that should be burned into your soul:**‘Complex roots of a polynomial equation with real coefficients form conjugate pairs’**.

⠀

This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and -1 – 2i are also roots of P(z) = 0. So we now have 4 roots of our**quartic **equation P(z) = 0, so that’s all of them!

⠀

We can now employ the**factor theorem**. Remember, this states that if z = α is a root of a polynomial then (z – α) is a factor of that polynomial.

⠀

So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that P(z) = (z – (2 - i))(z – (2 + i))(z – (-1 + 2i))(z – (-1 - 2i))

⠀

So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.

⠀

⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)

⠀

⇒ P(z) = (z² - 2z – zi – 2z + 4 + 2i + zi - 2i - i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i - 4i²)

⠀

⇒ P(z) = (z² - 4z + 5)(z² + 2z + 5) [Remember: i² = - 1]

⠀

⇒ P(z) = z⁴ + 2z³ + 5z² - 4z³ - 8z² - 20z + 5z² + 10z + 25

⠀

⇒ P(z) = z⁴ - 2z³ + 2z² - 10z + 25

⠀

Thus a = -2, b = 2, c = -10, d = 25 and we’re done.

⠀

Fact that should be burned into your soul:

⠀

This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and -1 – 2i are also roots of P(z) = 0. So we now have 4 roots of our

⠀

We can now employ the

⠀

So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that P(z) = (z – (2 - i))(z – (2 + i))(z – (-1 + 2i))(z – (-1 - 2i))

⠀

So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.

⠀

⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)

⠀

⇒ P(z) = (z² - 2z – zi – 2z + 4 + 2i + zi - 2i - i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i - 4i²)

⠀

⇒ P(z) = (z² - 4z + 5)(z² + 2z + 5) [Remember: i² = - 1]

⠀

⇒ P(z) = z⁴ + 2z³ + 5z² - 4z³ - 8z² - 20z + 5z² + 10z + 25

⠀

⇒ P(z) = z⁴ - 2z³ + 2z² - 10z + 25

⠀

Thus a = -2, b = 2, c = -10, d = 25 and we’re done.This is a question from an FP1 paper. Here i denotes √-1.

⠀

Fact that should be burned into your soul:

⠀

This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and -1 – 2i are also roots of P(z) = 0. So we now have 4 roots of our

⠀

We can now employ the

⠀

So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that P(z) = (z – (2 - i))(z – (2 + i))(z – (-1 + 2i))(z – (-1 - 2i))

⠀

So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.

⠀

⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)

⠀

⇒ P(z) = (z² - 2z – zi – 2z + 4 + 2i + zi - 2i - i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i - 4i²)

⠀

⇒ P(z) = (z² - 4z + 5)(z² + 2z + 5) [Remember: i² = - 1]

⠀

⇒ P(z) = z⁴ + 2z³ + 5z² - 4z³ - 8z² - 20z + 5z² + 10z + 25

⠀

⇒ P(z) = z⁴ - 2z³ + 2z² - 10z + 25

⠀

Thus a = -2, b = 2, c = -10, d = 25 and we’re done.

The first task is to translate the equation into something you’re more familiar with: write the equation in terms of sin(θ) and cos(θ). To do this use we’ll first need the definition of cosec(θ) and cot(θ). A trick for remembering which is which is to look at the third letter:

⠀

co__s__ec(θ) = 1/__s__in(θ) se__c__(θ) = 1/__c__os(θ) co__t__(θ) = 1/__t__an(θ)

⠀

So let’s use these definitions:

⠀

cosec(θ) + 5cot(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5/tan(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5cos(θ)/sin(θ) = 3sin(θ)

⠀

⇒ 1 + 5cos(θ) = 3sin²(θ)

⠀

Notice that we can now completely remove the sin(θ) term from the equation and leave only cos(θ) by using the identity sin²(θ) + cos²(θ) ≡ 1

⠀

⇒ 1 + 5cos(θ) = 3(1- cos²(θ))

⠀

⇒ 3cos²(θ) + 5cos(θ) - 2 = 0

⠀

That’s much nicer. We now have a quadratic equation in terms of cos(θ). Factorising gives (3cos(θ) - 1)(cos(θ) + 2) = 0 giving us solutions cos(θ) = 1/3 and cos(θ) = -2. Note that the second solution is not possible, since the graph of cos(θ) is bounded below by -1, so never attains the value -2. So we may discard it leaving only cos(θ) = 1/3 to worry about.

⠀

Plugging θ = arccos(1/3) into your calculator (with your calculator in radians!) gives 1.23 to 2 decimal places. However, this is only half the answer. The question asked for solutions in the** range** **0 to 2π**. We now have to use the cos(θ) graph to find all remaining solutions. I usually do this by drawing a quick sketch of cos(θ) from 0 to 2π and a horizontal line at y = 1/3.

⠀

Upon doing so you will see there are 2 solutions in this range. By the sketch and the symmetry of the cos(θ) graph you can see the other solution is 2π - 1.23 = 5.05 to 2 decimal places. As usual, plugging the solutions you get back into the equation is a great way to check whether you’ve got the correct answer!The first task is to translate the equation into something you’re more familiar with: write the equation in terms of sin(θ) and cos(θ). To do this use we’ll first need the definition of cosec(θ) and cot(θ). A trick for remembering which is which is to look at the third letter:

⠀

co__s__ec(θ) = 1/__s__in(θ) se__c__(θ) = 1/__c__os(θ) co__t__(θ) = 1/__t__an(θ)

⠀

So let’s use these definitions:

⠀

cosec(θ) + 5cot(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5/tan(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5cos(θ)/sin(θ) = 3sin(θ)

⠀

⇒ 1 + 5cos(θ) = 3sin²(θ)

⠀

Notice that we can now completely remove the sin(θ) term from the equation and leave only cos(θ) by using the identity sin²(θ) + cos²(θ) ≡ 1

⠀

⇒ 1 + 5cos(θ) = 3(1- cos²(θ))

⠀

⇒ 3cos²(θ) + 5cos(θ) - 2 = 0

⠀

That’s much nicer. We now have a quadratic equation in terms of cos(θ). Factorising gives (3cos(θ) - 1)(cos(θ) + 2) = 0 giving us solutions cos(θ) = 1/3 and cos(θ) = -2. Note that the second solution is not possible, since the graph of cos(θ) is bounded below by -1, so never attains the value -2. So we may discard it leaving only cos(θ) = 1/3 to worry about.

⠀

Plugging θ = arccos(1/3) into your calculator (with your calculator in radians!) gives 1.23 to 2 decimal places. However, this is only half the answer. The question asked for solutions in the** range** **0 to 2π**. We now have to use the cos(θ) graph to find all remaining solutions. I usually do this by drawing a quick sketch of cos(θ) from 0 to 2π and a horizontal line at y = 1/3.

⠀

Upon doing so you will see there are 2 solutions in this range. By the sketch and the symmetry of the cos(θ) graph you can see the other solution is 2π - 1.23 = 5.05 to 2 decimal places. As usual, plugging the solutions you get back into the equation is a great way to check whether you’ve got the correct answer!

⠀

co

⠀

So let’s use these definitions:

⠀

cosec(θ) + 5cot(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5/tan(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5cos(θ)/sin(θ) = 3sin(θ)

⠀

⇒ 1 + 5cos(θ) = 3sin²(θ)

⠀

Notice that we can now completely remove the sin(θ) term from the equation and leave only cos(θ) by using the identity sin²(θ) + cos²(θ) ≡ 1

⠀

⇒ 1 + 5cos(θ) = 3(1- cos²(θ))

⠀

⇒ 3cos²(θ) + 5cos(θ) - 2 = 0

⠀

That’s much nicer. We now have a quadratic equation in terms of cos(θ). Factorising gives (3cos(θ) - 1)(cos(θ) + 2) = 0 giving us solutions cos(θ) = 1/3 and cos(θ) = -2. Note that the second solution is not possible, since the graph of cos(θ) is bounded below by -1, so never attains the value -2. So we may discard it leaving only cos(θ) = 1/3 to worry about.

⠀

Plugging θ = arccos(1/3) into your calculator (with your calculator in radians!) gives 1.23 to 2 decimal places. However, this is only half the answer. The question asked for solutions in the

⠀

Upon doing so you will see there are 2 solutions in this range. By the sketch and the symmetry of the cos(θ) graph you can see the other solution is 2π - 1.23 = 5.05 to 2 decimal places. As usual, plugging the solutions you get back into the equation is a great way to check whether you’ve got the correct answer!The first task is to translate the equation into something you’re more familiar with: write the equation in terms of sin(θ) and cos(θ). To do this use we’ll first need the definition of cosec(θ) and cot(θ). A trick for remembering which is which is to look at the third letter:

⠀

co

⠀

So let’s use these definitions:

⠀

cosec(θ) + 5cot(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5/tan(θ) = 3sin(θ)

⠀

⇒ 1/sin(θ) + 5cos(θ)/sin(θ) = 3sin(θ)

⠀

⇒ 1 + 5cos(θ) = 3sin²(θ)

⠀

Notice that we can now completely remove the sin(θ) term from the equation and leave only cos(θ) by using the identity sin²(θ) + cos²(θ) ≡ 1

⠀

⇒ 1 + 5cos(θ) = 3(1- cos²(θ))

⠀

⇒ 3cos²(θ) + 5cos(θ) - 2 = 0

⠀

That’s much nicer. We now have a quadratic equation in terms of cos(θ). Factorising gives (3cos(θ) - 1)(cos(θ) + 2) = 0 giving us solutions cos(θ) = 1/3 and cos(θ) = -2. Note that the second solution is not possible, since the graph of cos(θ) is bounded below by -1, so never attains the value -2. So we may discard it leaving only cos(θ) = 1/3 to worry about.

⠀

Plugging θ = arccos(1/3) into your calculator (with your calculator in radians!) gives 1.23 to 2 decimal places. However, this is only half the answer. The question asked for solutions in the

⠀

Upon doing so you will see there are 2 solutions in this range. By the sketch and the symmetry of the cos(θ) graph you can see the other solution is 2π - 1.23 = 5.05 to 2 decimal places. As usual, plugging the solutions you get back into the equation is a great way to check whether you’ve got the correct answer!