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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: yasser37 on September 25, 2010, 08:13:35 am

Title: The Modulus
Post by: yasser37 on September 25, 2010, 08:13:35 am: Hi guys

can anyone gimme links that explains the mosulus to me please

I don't understand it from my teacher and I have a quiz tomorrow
Title: Re: The Modulus
Post by: Deadly_king on September 25, 2010, 08:37:12 am: Quote from: yasser37 on September 25, 2010, 08:13:35 am
Hi guys

can anyone gimme links that explains the mosulus to me please

I don't understand it from my teacher and I have a quiz tomorrow
well..........i dont know any links about that.

But what i can say for sure the modulus of something is always positive.

By the way which chapter are you dealing with??
Title: Re: The Modulus
Post by: elemis on September 25, 2010, 08:51:42 am: I am assuming you are referring to the modulus of a vector.

If $\begin{pmatrix}<br />2<br />\\ <br />1<br />\end{pmatrix}$ was the coordinates of a vector A.

Then A would be something like the diagram below.

The modulus helps us determine the magnitude of a vector. The modulus is calculated by : $\sqrt{x^2 + y^2}$

This is basically applying Pythagoras's Theorem to find the length of a hypotenuse.

NOTE : The modulus is ALWAYS POSITIVE. If you get a negative answer then you have done something wrong.
Title: Re: The Modulus
Post by: Arthur Bon Zavi on September 25, 2010, 09:17:02 am: Quote from: Ari Ben Canaan on September 25, 2010, 08:51:42 am

NOTE : The modulus is ALWAYS POSITIVE. If you get a negative answer then you have done something wrong.

If it is in negative, u can also find the determinant of it. ;D
Title: Re: The Modulus
Post by: yasser37 on September 25, 2010, 10:10:58 am: I am talking about chapter 2 in pure mathematics 2&3

for example in June 08 question 1

not sure how to do that
Title: Re: The Modulus
Post by: elemis on September 25, 2010, 10:35:46 am: $\frac{12}{sin45}*\frac{BC}{sin 60}$

Just solve for BC.
Title: Re: The Modulus
Post by: yasser37 on September 25, 2010, 10:52:27 am: June 08
Paper 3
Question 1

as an example

but I would really appreciate if I can find a link or something to explain the whole chapter 2 in Pure Mathematics 2&3
Title: Re: The Modulus
Post by: Deadly_king on September 25, 2010, 03:48:54 pm: Quote from: yasser37 on September 25, 2010, 10:52:27 am
June 08
Paper 3
Question 1

as an example

but I would really appreciate if I can find a link or something to explain the whole chapter 2 in Pure Mathematics 2&3
For this number, you should eliminate the modulus sign since the inequality sign will not agree for both positive and negative numbers. This can be done by squaring on both sides.

Let me get something clear for you.
The modulus sign is inserted to represent both the positive and negative values. Therefore it becomes complicated to solve this inequality if we use both the positive and negative values. Hence by squaring on both sides we are opting to work with the positive value only.

Example : (1)² = 1 but (-1)² is also equal to 1.

1. Squaring on both sides you'll obtain :
x² - 4x + 4 > 9(4x² +4x + 1)

Now you can solve the inequality as usual :)