Currently unavailable: for regular students
Degree: Mathematics (Bachelors) - Warwick University
Hi, I'm Jacob and I'm a second-year student of Mathematics at the University of Warwick. My main subject is Maths and I averaged 98.5% across my Maths and Further Maths A-level exams. I also maintain an interest in English and History and I'm enthusiastic to teach in these subjects.
About The Session
The structure of tutoring sessions will be flexible and tailored to the student. I will strive to help the student enjoy their subject but I also appreciate the necessity of preparing students for important exams.
I am both friendly and patient and will always try my best to answer students' questions no matter how big or small!
As well as the tutoring session, I will spend as much time as necessary outside of 'the classroom' to ensure that the tutoring sessions are as beneficial as possible for the student.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|STEP I||Uni Admissions Test||1|
There are two approaches to rationalising a denominator depending on how the denominator appears.
First we deal with the simpler case where the surd term is the only term in the denominator.
For example: 6/√3, 4/(3√7) or (5+√2)/√3
The trick of 'multiplying by one' can be used on each of these fractions to give an alternate expression that does not contain a surd in the denominator.
Let's consider the fraction 6/√3
If we multiply this fraction by √3/√3 we do not change the value of the fraction because √3/√3=1
Let's go ahead and do the multiplication:
The numerator is 6* √3 which is most simply expressed as 6√3
The denominator is √3*√3 which by the definition of the square root function is equal to 3
So putting the numerator and denominator together leaves us with the fraction (6√3)/3
Some basic cancellation reduces this to the simpler form of 2√3
Next comes the more complicated case where the surd term is not the only term in the denominator.
For example: 1/(1-√3), (1+√2)/(3+√2) or 6/(2√5+√3)
The trick of 'multiplying by one' can again be used but it is less clear what 'one' to use.
To find our 'one' we must recall the difference of two squares formula: a2-b2=(a+b)(a-b)
Let's consider 1/(1-√3)
From the difference of two squares formula we can see that if we multiply the denominator by 1+√3 then we are left with 12-(√3)2=1-3=-2
From this we can infer that it is necessary to multiply the fraction by (1+√3)/(1+√3)
Let's go ahead and do the multiplication:
The numerator is 1*(1+√3) which is equal to 1+√3
The denominator we have already calculated is -2
So putting the numerator and the denominator together leaves us with (1+√3)/-2
A useful reason for rationalising the denominator is that it helps when thinking about what value a fraction really represents. For example, when considering the fraction 1/√2=√2/2, it is hard to imagine 1 divided into √2 pieces as √2 is an irrational number. It makes more sense however to imagine √2 divided into 2 pieces as that is just a simple halving and it doesn't matter too much that √2 is irrational.see more
Every complex number has complex square roots. However since we don't know how to deal with expressions such as √i we need to follow a specific method to find the square roots of a complex number.
Let's consider the complex number 21-20i.
We know that all square roots of this number will satisfy the equation 21-20i=x2 by definition of a square root.
We also know that x can be expressed as a+bi (where a and b are real) since the square roots of a complex number are always complex.
The natural step to take here is the mulitply out the term on the right-hand side.
This gives 21-20i=a2+(2ab)i+(b2)i2.
As i2=-1 by definition of i, this equation can be rearranged to give 21-20i=(a2-b2)+(2ab)i.
Now both sides of the equation are in the same form.
Let's compare coeffiecients to obtain two equations in a and b.
First, let's compare the real parts of the equation.
We have a2-b2=21 (call this equation 1).
Next, let's compare the imaginary parts of the equation (the coefficients of i).
We have 2ab=-20 (call this equation 2).
We now have two equations in two unknowns. We can solve these simultaneous equations for a and b.
Firstly, we can make b the subject of equation 2 by dividing both sides by 2a.
We have b=-10/a.
Now substitute this expression for b into equation 1.
We have a2-(-10/a)2=21.
Some simplification and factorisation of this equation gives us (a2+4)(a2-25)=0, a quadratic in disguise.
So either a2=-4 or a2=25.
We have assumed a to be real so a2=-4 has no solutions of interest to us.
This means our solutions are a=5 and a=-5.
Substitute each a value into our earlier expression for b.
This means that when a=5, b=-2 and when a=-5, b=2.
So, putting a and b back into the context of the question, we have two solutions: 5-2i and -5+2i.see more