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

## 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)

jamesrc
Gold Member

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?

Fredrik
Staff Emeritus
Gold Member

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.

Mark44
Mentor

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?

mathman

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.