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

  • Thread starter vkash
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  • #1
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Main Question or Discussion Point

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)
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
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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?
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
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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.
 
  • #4
33,168
4,852


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?
 
  • #5
mathman
Science Advisor
7,758
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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.
 
  • #6
318
1


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

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