Hi, I'm Eden, a fourth year engineering student at the University of Oxford. When I was taking my GCSEs my Maths teacher showed me how maths is the language we use to describe our reality, and ever since then I've been impassioned to share this wondrous view through tutoring.

I have experience tutoring at all levels of maths and physics, so will guarantee you understand why things are the way they are and where they came from – not just how to do them. I’m patient and friendly, listening carefully to any problems you may have before helping you to logically think them through.

During the lessons we will move at a pace that you are comfortable with, whilst covering all of the necessary ground. I’m sure that by the end you will be able to explain the problem to me as well as I can explain it to you! My hope is that you will come to see maths not as the tedious subject that schoolteachers have a habit of making it out to be, but the gateway through which we can understand our universe!

If you’re interested, or have any questions, please send me a ‘WebMail’ or book a ‘Meet the Tutor Session’! (both accessible through this website).

I look forwards to meeting you!

Thanks, Eden

Hi, I'm Eden, a fourth year engineering student at the University of Oxford. When I was taking my GCSEs my Maths teacher showed me how maths is the language we use to describe our reality, and ever since then I've been impassioned to share this wondrous view through tutoring.

I have experience tutoring at all levels of maths and physics, so will guarantee you understand why things are the way they are and where they came from – not just how to do them. I’m patient and friendly, listening carefully to any problems you may have before helping you to logically think them through.

During the lessons we will move at a pace that you are comfortable with, whilst covering all of the necessary ground. I’m sure that by the end you will be able to explain the problem to me as well as I can explain it to you! My hope is that you will come to see maths not as the tedious subject that schoolteachers have a habit of making it out to be, but the gateway through which we can understand our universe!

If you’re interested, or have any questions, please send me a ‘WebMail’ or book a ‘Meet the Tutor Session’! (both accessible through this website).

I look forwards to meeting you!

Thanks, Eden

No DBS Check

First label all the sides with letters:

longest side = a

middle length side = b

shortest side = c

Now convert the question into equation form:

"the longest side is two units more than the shortest side" becomes, a = 2 + c (1)

"middle length side is one unit longer than the shortest side" becomes, b = 1 + c (2)

"The total surface area of the cuboid is 52 units² " becomes, 2*a*b + 2*a*c + 2*b*c = 52 (3)

Substitute expressions (1) and (2) into (3) to get

2(2+c)(1+c) + 2c(2+c) + 2c(1+c) = 52

Expand out the brackets:

2(2+3c+c^2) + 4c + 2c^2 + 2c + 2c^2 = 52

which becomes:

4 + 6c + 2c^2 + 6c + 4c^2 = 52

collect all the terms together:

6c^2 + 12c - 48 = 0

Divide by 6 to simplify equation:

c^2 + 2c - 8 = 0

factorise this equation into brackets to find c, by finding two numbers that multiply together to make -8, and add together to make 2... i.e. 4 and -2,

(c + 4)(c-2) = 0

Therefore, c = -4, or c=2. Since c must be positive (a negative side length would make no physical sense) c must be 2. We can then subsitute this back into (1) and (2) to find values for a and b,

a = 4

b = 3

and then double check that the surface area is 52.

First label all the sides with letters:

longest side = a

middle length side = b

shortest side = c

Now convert the question into equation form:

"the longest side is two units more than the shortest side" becomes, a = 2 + c (1)

"middle length side is one unit longer than the shortest side" becomes, b = 1 + c (2)

"The total surface area of the cuboid is 52 units² " becomes, 2*a*b + 2*a*c + 2*b*c = 52 (3)

Substitute expressions (1) and (2) into (3) to get

2(2+c)(1+c) + 2c(2+c) + 2c(1+c) = 52

Expand out the brackets:

2(2+3c+c^2) + 4c + 2c^2 + 2c + 2c^2 = 52

which becomes:

4 + 6c + 2c^2 + 6c + 4c^2 = 52

collect all the terms together:

6c^2 + 12c - 48 = 0

Divide by 6 to simplify equation:

c^2 + 2c - 8 = 0

factorise this equation into brackets to find c, by finding two numbers that multiply together to make -8, and add together to make 2... i.e. 4 and -2,

(c + 4)(c-2) = 0

Therefore, c = -4, or c=2. Since c must be positive (a negative side length would make no physical sense) c must be 2. We can then subsitute this back into (1) and (2) to find values for a and b,

a = 4

b = 3

and then double check that the surface area is 52.

If the roots of this cubic equation are a, b and c, then the equation can be written

(z - a)(z - b)(z - c) = 0

multiplying this out gives:

z^3 - az^2 - bz^2 - cz^2 + abz + acz + bcz - abc

grouping terms with z to the same power in them, this becomes:

z^3 + (-a - b - c)z^2 + (ab + ac + bc)z - abc

Equating the coefficients in the equation above with the coefficients of the equation in the question gives the following equations:

(-a -b - c) =2

or

a + b + c = -2

and

(ab + ac + bc) = 3

and

-abc = -4.

To show that a^2 + b^2 + c^2 = -2, first we need to manipulate our expressions above to get an expression with a^2 + b^2 +c^2 in it. The most obvious way to do this is by squaring (a + b +c) i.e.

(a + b + c)^2 = (-2)^2

which multiplies out to give

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 4

-> a^2 + b^2 + c^2 + 2(ab + ac + bc) = 4

We now that (ab + ac + bc) = 3, so substitue this into the expression above

a^2 + b^2 + c^2 + 2(3) = 4

Finally, take away 6 from both sides to get,

a^2 + b^2 + c^2 = -2

If the roots of this cubic equation are a, b and c, then the equation can be written

(z - a)(z - b)(z - c) = 0

multiplying this out gives:

z^3 - az^2 - bz^2 - cz^2 + abz + acz + bcz - abc

grouping terms with z to the same power in them, this becomes:

z^3 + (-a - b - c)z^2 + (ab + ac + bc)z - abc

Equating the coefficients in the equation above with the coefficients of the equation in the question gives the following equations:

(-a -b - c) =2

or

a + b + c = -2

and

(ab + ac + bc) = 3

and

-abc = -4.

To show that a^2 + b^2 + c^2 = -2, first we need to manipulate our expressions above to get an expression with a^2 + b^2 +c^2 in it. The most obvious way to do this is by squaring (a + b +c) i.e.

(a + b + c)^2 = (-2)^2

which multiplies out to give

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 4

-> a^2 + b^2 + c^2 + 2(ab + ac + bc) = 4

We now that (ab + ac + bc) = 3, so substitue this into the expression above

a^2 + b^2 + c^2 + 2(3) = 4

Finally, take away 6 from both sides to get,

a^2 + b^2 + c^2 = -2

The standard rule for integration is: integrate kx^n dx = kx^(n+1)/(n+1). However, if we try and integrate 1/x in this manner we get, x^0/0, i.e. 1/0, which is infinity. However, if we look at a graph of 1/x, then between two points there is clearly a well defined area, so it must be possible to integrate this. The natural logarithm is a function that we use to do this, whereby ln(a) is the integral of 1/x between 1 and a. It is a logarithmic function with base 'e', where e takes the value of about 2.718, and e^x is known as the exponential function; i.e. it increases at an ever increasing rate. The exponential function is the inverse of the natural logarithm function.

The standard rule for integration is: integrate kx^n dx = kx^(n+1)/(n+1). However, if we try and integrate 1/x in this manner we get, x^0/0, i.e. 1/0, which is infinity. However, if we look at a graph of 1/x, then between two points there is clearly a well defined area, so it must be possible to integrate this. The natural logarithm is a function that we use to do this, whereby ln(a) is the integral of 1/x between 1 and a. It is a logarithmic function with base 'e', where e takes the value of about 2.718, and e^x is known as the exponential function; i.e. it increases at an ever increasing rate. The exponential function is the inverse of the natural logarithm function.