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modulus

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modulus [#permalink] New post 11 Apr 2014, 05:30
solution for x-|2x+1|=3?
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Re: modulus [#permalink] New post 18 Apr 2014, 08:03
after removing the modulus we get two values of x and neither of them satisfy the equation so there be no solution.
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Re: modulus [#permalink] New post 20 May 2014, 08:46
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Dear Shreshth

As you have well noticed, when you open the modulus and solve the equation, you get two values: x= 2/3 and x= -4. None of these however satisfy the equation!

I'll explain why this happens.

We can rewrite the given equation as:

|2x+1| = x-3.

I've drawn the graph of y= |2x+1| below.

Attachment:
Mod Ques.PNG
Mod Ques.PNG [ 9.72 KiB | Viewed 277 times ]


As becomes clear from the graph,

For x>-1/2, |2x+1| = 2x+1
For x< -1/2, |2x+1| = -(2x+1)


So, when we open the modulus, we should actually write the equations like this:

For x<-1/2,

-(2x+1)= x-3
-2x-x= -3+1
-3x= -2
x= 2/3

However, since this solution falls outside the region x<-1/2 for which the equation is valid, it cannot be accepted.

Similarly,

For x >1/2,

2x+1 = x-3
2x-x= -3-1
x= -4

However, since this solution falls outside the region x>1/2, for which the equation is valid, it cannot be accepted.
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Re: modulus [#permalink] New post 20 May 2014, 08:55
The above question very effectively teaches us that we should be careful when opening the modulus.

When you open the modulus |ax+b| as (ax+b), x should be greater than -b/a. Do check that your solution for the equation that uses (ax+b) does indeed fall in this region.

And, when you open the modulus |ax+b| as -(ax+b), x should be lesser than -b/a.

Thanks Shreshth for sharing this instructive question!
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Re: modulus   [#permalink] 20 May 2014, 08:55
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