Henry L. IB Maths tutor, A Level Maths tutor, 13 plus  Maths tutor, 1...

Henry L.

Currently unavailable: for new students

Degree: Chemical Engineering (Masters) - Bath University

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About me

About me

I am a chemical engineering student at the University of Bath. Mathematics and the sciences are subjects I have always enjoyed and I especially love applying the knowledge I have learnt in all sorts of areas, such as cooking or music or difficult engineering problems. I teach my younger brother (age 9) mathematics and he is currently working at a year 7 level. Throughout sixth form I also spent time teaching younger children science and so I am fairly experienced at teaching.

I hope to make my tutorials exciting and interesting for you and to show you that there is fun in learning and that it should not be just for passing an exam, but I will prioritise helping you get the highest grade you can.  When I started IB HL Maths I was scoring 4s and 5s despite achieving the highest grades in IGCSE Maths and Admaths but I ultimately finished with a 7. Therefore, I know full well how difficult mathematics can be and I am very patient and friendly when it comes to teaching.

Sessions

A strong foundation is essential for learning. Mathematics is a cumulative subject and you must understand certain ideas before progressing. This is something I did not appreciate and after I learned of the importance of understanding I found maths a lot easier and more enjoyable. I therefore would like to focus on helping you develop your understanding of topics. This will allow you to not only complete ‘standard’ exam questions but to tackle any ‘unusual’ questions exams may give.

Personal Statements

These can be stressful and painful to write but when it came to writing my own I found it to be the opposite. I’ve done a lot of research into what university admissions are looking for and hope that you will find my advice helpful. When I applied, it only took one small redraft before my tutors and I were happy and confident in my personal statement, and I was ultimately invited to an interview at Cambridge.

Tutoring

Please let me know before sessions what you would like to cover, and how much of the topic(s) in question you understand. I ask this so I can prepare well for sessions to help you get the most out of them. 

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr
Maths IB £20 /hr
Maths 13 Plus £18 /hr
Maths 11 Plus £18 /hr
-Personal Statements- Mentoring £20 /hr

Qualifications

QualificationLevelGrade
IB Higher Level MathematicsBaccalaureate7
IB Higher Level PhysicsBaccalaureate7
IB Higher Level ChemistryBaccalaureate7
IB Standard Level EconomicsBaccalaureate7
IB Standard Level EnglishBaccalaureate6
IB Standard Level LatinBaccalaureate6
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

Currently unavailable: for new students

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Ratings and reviews

5from 22 customer reviews

Elle (Student) September 18 2016

great help

Elle (Student) August 28 2016

very helpful

Zoe (Parent) July 18 2016

So helpful thank you

Elle (Student) July 11 2016

very helpful
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Questions Henry has answered

Factorise z^3+1 into a linear and quadratic factor. Let y=(1+i√3)/2. Show that y is a cube root of -1. Show that y^2=y-1. Find the value of (1-y)^6.

To factorise z3+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z3+1=0. We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)3+1=0 since (-1)3=-1. There...

To factorise z3+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z3+1=0.

We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)3+1=0 since (-1)3=-1.

Therefore, (z+1) is a factor. The other factor is quadratic. We know, by inspection, that the quadratic factor will look like: (z2+kx+1) where k is a constant. This is because z3+1 can be written as a product of its factors: z3+1=(z+1)(z2+kx+1). The coefficient of the z3 term and constant term are 1 so we know the coefficient of the z2 and constant term in the quadratic. By inspection we can also see that k=-1.

The second part of the questions asks us to show that y3=-1. There are several approaches for this. You can square y to obtain y2 then multiply it by y again to obtain y3 and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y3 then convert back into Cartesian form. The final method is as follows:

We want to show that y3=-1. Earlier we wrote: z3=-1 which leads to z3+1=(z+1)(z2-z+1)=0. We know that if y is a cubic root of -1 then it cannot be equal to -1, so for the above to be true then (z2-z+1) must be equal to 0. We simply solve the quadratic (z2-z+1)=0 using the quadratic formula. The formula gives us that z=(1+i√3)/2 after taking the positive result. We note that this is equal to y. Since y=x then y is a cube root of -1.

The third part of the question asks us to show that y2=y-1. Using our working from earlier, we know that y is a root of z2-z+1=0 so y2-y+1=0. Rearranging this we have y2=y-1 as required. Alternatively you can find y2  then show that it is equal to y-1.

The final part of the question is to find the value of (1-y)6. Instinctively most people would use the binomial expansion but there is a far more elegant solution. From the previous part we have y2=y-1. If we multiply both sides by -1 then:

(-y2 )=1-y

Raising both sides to the power 6 as the question required and using power laws:

(-y2)6=(1-y)6

Why would we do this though? Maybe you can see why; there is a piece of information from earlier that we can use, if not, keep following and you will see that this method is very clever.

(-y)6*2=(y)12=(1-y)6

The negative sign disappears as squaring a number makes it positive.

We can manipulate this further:

(y)12=(y)3*4=(y3)4

Earlier we had (y3)=-1 and so ultimately the value of (1-y)6=(-1)4=1.

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1 year ago

1407 views

Make y the subject of (y/x)+(2y/(x+4))=3

We want to make y the subject of the equation and so we need to write it in the form y=f(x) where f(x) is a function in terms of x. When rearranging equations with fractions for a certain subject, factorising will usually be involved. A good start to this would be to put everything over a com...

We want to make y the subject of the equation and so we need to write it in the form y=f(x) where f(x) is a function in terms of x.

When rearranging equations with fractions for a certain subject, factorising will usually be involved. A good start to this would be to put everything over a common denominator (x(x+4)) which is the product of the two different denominators in the original equation. Recall that to maintain equality (i.e. the left hand side equals the right hand side), what you do to one side you must also do to the other side. This is an incredibly important rule that will help you solve the majority, if not, all of maths problems you will come across while avoiding mistakes.

So, to (y/x), we multiply by (x+4)/(x+4). Note that this is equal to 1, as any number divided by itself is 1, and any number multipled by 1 is itself so we aren't actually changing the equation. Simlarly for (2y/(x+4)) we multiply by (x/x). For the right hand side of the equation, we can multiply by (x(x+4))/(x(x+4)).

This gives us:

(y(x+4)+2xy)/(x(x+4))=3(x(x+4))/(x(x+4))

At this point we can multiply everything on both sides of the equation by (x(x+4)) to remove the denominator. We can do this because of the rule mentioned earlier: what we do to one side of the equation we must also do to the other side of the equation.

After this step we have:

xy+4y+2xy=3(x(x+4))

We want y as the subject of the equation so on the left hand side we factorise out the y term, as every term on the left hand side has y in it.

y(x+4+2x)=y(4+3x)=3(x(x+4))

Finally, we divide both sides by (4+3x):

y=(x(x+4))/(4+3x)

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1 year ago

905 views
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