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Degree: Mathematics (Bachelors)  Warwick University
About Me
Hi, I'm Jacob and I'm a secondyear student of Mathematics at the University of Warwick. My main subject is Maths and I averaged 98.5% across my Maths and Further Maths Alevel 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.
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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/âˆš3, 4/(3âˆš7) or (5+âˆš2)/âˆš3
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/âˆš3
If we multiply this fraction byÂ âˆš3/âˆš3 we do not change the value of the fraction becauseÂ âˆš3/âˆš3=1
Let's go ahead and do the multiplication:
The numerator is 6*Â âˆš3 which is most simply expressed as 6âˆš3
The denominator is âˆš3*âˆš3 which by the definition of the square root function is equal to 3
SoÂ putting the numerator and denominator togetherÂ leaves us with the fraction (6âˆš3)/3
Some basic cancellation reduces this to the simpler form of 2âˆš3
So 6/âˆš3=2âˆš3
Next comes the more complicated case where the surd term is not the only term in the denominator.
For example: 1/(1âˆš3), (1+âˆš2)/(3+âˆš2) or 6/(2âˆš5+âˆš3)
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: a^{2}b^{2}=(a+b)(ab)
Let's consider 1/(1âˆš3)
From the difference of two squares formula we can see that if we multiply the denominator by 1+âˆš3 then we are left with 1^{2}(âˆš3)^{2}=13=2
From this we can infer that it is necessary to multiply the fraction by (1+âˆš3)/(1+âˆš3)
Let's go ahead and do the multiplication:
The numerator is 1*(1+âˆš3) which is equal to 1+âˆš3
The denominator we have already calculated is 2
So putting the numerator and the denominator together leaves us with (1+âˆš3)/2
So 1/(1âˆš3)=(1+âˆš3)/2
A useful reason for rationalising the denominator is thatÂ it helps when thinking about what value a fraction really represents. For example, when considering the fraction 1/âˆš2=âˆš2/2, it is hard to imagine 1 divided intoÂ âˆš2 pieces asÂ âˆš2 is an irrational number. It makes more sense however to imagineÂ âˆš2 divided into 2 pieces as that is just a simple halving and it doesn't matter too much thatÂ âˆš2 is irrational.
see moreEvery complex number has complex square roots. However since we don't know how to deal with expressions such asÂ âˆši we need to follow a specific method to find the square roots of a complex number.
Let's consider the complex number 2120i.
We know that all square roots of this number will satisfy the equation 2120i=x^{2}Â 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.
So 2120i=(a+bi)^{2}.
The natural step to take here is the mulitply out the term on the righthand side.
This gives 2120i=a^{2}+(2ab)i+(b^{2})i^{2}.
As i^{2}=1 by definition of i, this equation can be rearranged to give 2120i=(a^{2}b^{2})+(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 a^{2}b^{2}=21 (call this equation 1).
Next, let's compare the imaginary parts of theÂ 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 a^{2}(10/a)^{2}=21.
Some simplification and factorisation of this equation gives us (a^{2}+4)(a^{2}25)=0, a quadratic in disguise.
So either a^{2}=4 or a^{2}=25.
We have assumed a to be real so a^{2}=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:Â 52i and 5+2i.
see more