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20 May 2014, 09:46
Kudos
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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.
Mod Ques.PNG [ 9.72 KiB | Viewed 2074 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.
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 [ 9.72 KiB | Viewed 2074 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.
