Confused with the answer<> seems correct buttht's wrong wrong?

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The discussion centers on solving the equation sqrt(x+1) - sqrt(x-1) = sqrt(4x-1). After squaring both sides and rearranging, the solution x = 5/4 is found, but it does not satisfy the original equation. Participants clarify that squaring both sides can introduce extraneous solutions, leading to incorrect conclusions about the existence of real solutions. The error arises from misunderstanding how squaring affects the equality, as it can yield additional, invalid solutions. Ultimately, the key takeaway is that squaring both sides of an equation can complicate the solution process by introducing false positives.
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confused with the answer<> seems correct buttht's wrong wrong??

question is
find solution
sqrt(x+1)-sqrt(x-1)=sqrt(4x-1)
sqrt(x+1)-sqrt(x-1)=sqrt(4x-1) - - - - - - - - - - - - - - - - - - - - - - - (1)
squaring both sides
(x+1)+(x-1)-2*sqrt(x2-1)=4x-1 - - - - - - - - - - - - - - - - (2)
solving and rearranging
1-2x=2*sqrt(x2-1) - - - - - - - - - - - - - - - - - - - - - - - -(3)
once again squaring both sides;
1-4x= -4
x=5/4;
But it does not satisfy the first equation.
it also doesn't satisfying equation number three, Is it reason for this?
If yes then why it is so?>?>?>?>?>?>(this is my question)
 
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Are you sure that your solution doesn't satisfy those equations? When you take the square root of a number, how many solutions do you get?
 


You seem to have started with an equation that doesn't have any real solutions. Let's consider a simpler problem: Find all real numbers x such that ##\sqrt x =-1##. If you square both sides, you get x=1. But x=1 doesn't satisfy the original equation, since ##\sqrt 1=1\neq -1##.

By squaring both sides, we only proved that if ##\sqrt x=-1##, then ##x=1##. This is an implication, not an equivalence, since x=1 doesn't imply ##\sqrt x=-1##. So we can't conclude that x=1. We can only conclude that there are no solutions with x≠1.
 


jamesrc said:
Are you sure that your solution doesn't satisfy those equations? When you take the square root of a number, how many solutions do you get?
I'm not sure where you're going with this question.

When you take the square root of a number, you get one value. Were you going to suggest that there are two?
 


vkash said:
question is

sqrt(x+1)-sqrt(x-1)=sqrt(4x-1) - - - - - - - - - - - - - - - - - - - - - - - (1)
squaring both sides
(x+1)+(x-1)-2*sqrt(x2-1)=4x-1 - - - - - - - - - - - - - - - - (2)
solving and rearranging
1-2x=2*sqrt(x2-1) - - - - - - - - - - - - - - - - - - - - - - - -(3)
once again squaring both sides;
1-4x= -4
x=5/4;
But it does not satisfy the first equation.
it also doesn't satisfying equation number three, Is it reason for this?
If yes then why it is so?>?>?>?>?>?>(this is my question)
Equation (3) lhs = -3/2, rhs = 3/2, so the squares are =, which is the source of your problem.
 


thanks to all of you;
i have got the point of error.
squaring add some extra answers to our solutions...
 
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