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Old 17th April 2012, 07:36 PM   #1 (permalink)
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AM Modulus question

I have a question regarding the steps to solve this |x| + |x+2| = 3 (ans x= -2.5, 0.5)
I did the first method which was to use the concept of |x| = a, therefore x = a or x = -a.
but the second method was to consider 3 regions:
x < -2, -2 <= x < 0, x >= 0
and that was what I don't really understand. My teacher said it was the JC way of doing it, can somebody explain it to me?

Thanks!

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Old 17th April 2012, 08:50 PM   #2 (permalink)
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Re: AM Modulus question

Your description is sorta vague but what I think is meant is the following.

If x<-2, then -x-(x+2)=3 => -2x-2=3 => x=-2.5

If -2≤x<0, then -x+x+2=3 => no solution

If 0≤x, then x+x+2=3 => x=0.5

I'll leave you to analyse the solution first, then feel free to ask if you wish


Last edited by Betakuwe; 17th April 2012 at 08:51 PM.
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Old 17th April 2012, 10:49 PM   #3 (permalink)
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Re: AM Modulus question

Yeah. That's the part where I don't get it.
How do you even get the 3 regions? Andhow do you conclude that if x < -2, then -x - (x+2)...?

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Old 17th April 2012, 11:29 PM   #4 (permalink)
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Re: AM Modulus question

Okay firstly you are finding values of x between -∞ and ∞ that satisfy the equation right?

Then we consider 3 different types of x, namely, x<-2, -2≤x<0, and 0≤x. These three cover all x's between -∞ and ∞.

If x<-2, then x and x+2 will ALWAYS be negative right? So if x<-2, we can remove the absolute value signs and negate x and x+2, that will make them back to positive. Therefore -x-(x+2)=3 if x<-2.

Then if -2≤x<0, x will be ALWAYS be negative and x+2 will ALWAYS be positive. So we can remove the absolute value signs and negate x so that it will be positive. Hence, -x+x+2=3 if -2≤x<0.

Lastly if x≥0, then both x and x+2 will ALWAYS be positive right? So the absolute value signs can just be taken away since both of them are already positive. Thus, x+x+2=3 if x≥0.

Note: When I say "positive" I actually meant both positive and zero, |0|=0 so yeah. Zero is not positive btw, lol.


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Old 26th April 2012, 09:46 AM   #5 (permalink)
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Re: AM Modulus question

Originally Posted by JiaJian View Post
I have a question regarding the steps to solve this |x| + |x+2| = 3 (ans x= -2.5, 0.5)
I did the first method which was to use the concept of |x| = a, therefore x = a or x = -a.
but the second method was to consider 3 regions:
x < -2, -2 <= x < 0, x >= 0
and that was what I don't really understand. My teacher said it was the JC way of doing it, can somebody explain it to me?

Thanks!
If this is the AM questions, den i try using the Amath method which should be much longer but within ur knowledge.

Knowing that to remove the modulus, one way is to square it. I.e. |x| = a => x^2 = a^2.
we have.

|x| + |x+2| = 3
By squaring both side,
x^2 + 2|x||x+2| + (x+2)^2 = 9
2|x||x+2| = 9 - x^2 - (x+2)^2
2|x||x+2| = 9 - x^2 - x^2 - 4x - 4
2|x||x+2| = -(2x^2 + 4x +5)
By squaring both side again,
4(x^2)(x+2)^2 = (2x^2 + 4x +5)^2
well, expand both side, simplify it and you may be able to solve.

Above are the standard method of solving the stated questions for Amath. As you can see JC methods are simpler.

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Old 26th April 2012, 09:51 AM   #6 (permalink)
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Re: AM Modulus question

How do you get the 3 regions:

1) Consider |x|. So it's either x>0 or x<0.

2) Consider |x+2|. So it's either x+2>0 or x+2<0, ie. x>-2 or x<-2

3) You "combine" the inequalities..

x<0 and x<-2 will give you the first region of x<-2

x<0 and x>-2 will give you the second region of -2<x<0

x>0 and x>-2 will give you the third region of x>0

Hence, you have the 3 regions to consider.

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