Andrew P. A Level Maths tutor, GCSE Maths tutor

Andrew P.

£18 - £20 /hr

Currently unavailable: for regular students

Studying: Mathematics (Bachelors) - Birmingham University

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About me

I am a Mathematics student at Birmingham University. I think Maths is a great subject with countless real life applications, and I love helping other people understand and explore it!

Passing on my knowledge and skills to others is an area I'm very comfortable with, as I have a year's Maths and English tutoring experience with Kumon. Additionally I was a youth basketball coach for 4 years before I moved to university.

I regard myself as a friendly and approachable person, so you can expect our sessions to have lots of discussion. I believe that the most effective and interesting sessions are led by the student, with the tutor acting as a guide. In the end, it's all about you and developing your knowledge!

Please feel free to contact me if you have any questions! I hope to speak to you soon.

I am a Mathematics student at Birmingham University. I think Maths is a great subject with countless real life applications, and I love helping other people understand and explore it!

Passing on my knowledge and skills to others is an area I'm very comfortable with, as I have a year's Maths and English tutoring experience with Kumon. Additionally I was a youth basketball coach for 4 years before I moved to university.

I regard myself as a friendly and approachable person, so you can expect our sessions to have lots of discussion. I believe that the most effective and interesting sessions are led by the student, with the tutor acting as a guide. In the end, it's all about you and developing your knowledge!

Please feel free to contact me if you have any questions! I hope to speak to you soon.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Physical EducationA-level (A2)A*
BiologyA-level (A2)A

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions Andrew has answered

Solve the quadratic equation x^2 + 4x +1 = 0 by completing the square.

Completing the square means to put our equation into a slightly different form which looks like this, where a and b are real numbers:

(x+a)2 + b = 0

From here, we can rearrange the equation and directly solve for x. Let's have a look at our specific example:

x2 +4x +1 = 0

The first step is to divide the coefficient of x by 2, and add this to x (this is our value of 'a' to go inside our bracket). We then square this value of a and subtract it outside the bracket.

In our example it will look like this:

(x+2)2 - 4 + 1 = 0

(x+2)2 - 3 = 0

We have our equation in completed square form.

[There is a quick way to check we've got this right by expanding out this equation quickly:

(x+2)(x+2) - 3 = 0

x2 + 4x + 4 - 3 = 0

x2 + 4x +1 = 0

We're back to our original equation, so we know we've got it right. Let's go and solve our equation in completed square form.]

We simply rearrange for x:

(x+2)2 - 3 = 0

Add 3 to both sides.

(x+2)2 = 3

Take the square root of both sides. This splits into two possible cases:

Case 1: Positive square root of 3

x+2 = + sqrt(3)

x = - 2 + sqrt(3)

Case 2: Negative square root of 3

x+2 = - sqrt(3)

x = - 2 - sqrt(3)

So our final answer is...

x = - 2 + sqrt(3)

x = - 2 - sqrt(3)

Completing the square means to put our equation into a slightly different form which looks like this, where a and b are real numbers:

(x+a)2 + b = 0

From here, we can rearrange the equation and directly solve for x. Let's have a look at our specific example:

x2 +4x +1 = 0

The first step is to divide the coefficient of x by 2, and add this to x (this is our value of 'a' to go inside our bracket). We then square this value of a and subtract it outside the bracket.

In our example it will look like this:

(x+2)2 - 4 + 1 = 0

(x+2)2 - 3 = 0

We have our equation in completed square form.

[There is a quick way to check we've got this right by expanding out this equation quickly:

(x+2)(x+2) - 3 = 0

x2 + 4x + 4 - 3 = 0

x2 + 4x +1 = 0

We're back to our original equation, so we know we've got it right. Let's go and solve our equation in completed square form.]

We simply rearrange for x:

(x+2)2 - 3 = 0

Add 3 to both sides.

(x+2)2 = 3

Take the square root of both sides. This splits into two possible cases:

Case 1: Positive square root of 3

x+2 = + sqrt(3)

x = - 2 + sqrt(3)

Case 2: Negative square root of 3

x+2 = - sqrt(3)

x = - 2 - sqrt(3)

So our final answer is...

x = - 2 + sqrt(3)

x = - 2 - sqrt(3)

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2 years ago

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