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# Modulus? :/ Watch

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1. Hi guys,

There is nothing in any of my books about modulus :/
This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

|2x-4|=|3x+2|

Thanks

2. So

However, it's unlikely all the solutions to will satisfy

So you'll need to substitute any found values of x back into the above equation to ensure it's satisfied.
3. (Original post by Noble.)

So

I'm more confused than before
4. (Original post by dream123)
Hi guys,

There is nothing in any of my books about modulus :/
This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

|2x-4|=|3x+2|

Thanks
You need to think about the regions where:
(i) and
(ii) and
(iii) and

Then use the fact that to solve it. Bear in mind, though, that (i) and (iii) will actually give you the same answer.
5. (Original post by dream123)
Hi guys,

There is nothing in any of my books about modulus :/
This isn't the question that I need to answer-I've made it up to use as a 'template' to help answer the actual question-any help would be appreciated!

|2x-4|=|3x+2|

Thanks
Because mod f(x) is always positive, you need to add a function on both sides that makes it positive. Squaring both sides is normally used

becomes

NOTE: This only works if one side is purely a modulus, mixing scalars and modulus isn't as easy.
6. (Original post by dream123)
I'm more confused than before
Basically, when you take the modulus of an entire function, you know it cannot exist below the y axis. It's reflected on the x-axis. Graphically, when you reflect this function, you'll always go from a positive to negative gradient, or from a negative to a positive gradient. Which is essence creates two functions. The original function, and another function with a gradient of the opposite sign.

Because the idea behind the modulus sign is to just take the magnitude of a function, irrespective of it's sign, you know that

and

and so

So when you consider , the modulus of that f(x) produces the two functions.

when considering x values that are

when considering x values that are

Because the function hits the x-axis at x=2, so it's the point at which the function is reflected. http://www.wolframalpha.com/input/?i=|2x-4|

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Updated: December 2, 2010
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