There are several typical mistakes people make on these, so we can't tell which one you're making without seeing your work (in some detail).Hello guys, I just started studying irrational functions and I'm stuck at this equation. The solutions say the result is X=1 but I can't seem to get there.
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Following Jomo's advice:Hello guys, I just started studying irrational functions and I'm stuck at this equation. The solutions say the result is X=1 but I can't seem to get there.
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Simply brilliant!Just out of interest
[MATH]\sqrt{2x-1}=1-\sqrt{x-1}[/MATH]
[MATH]\sqrt{x-1}[/MATH] shows that [MATH]\boxed{\;x≥1\;}[/MATH]
[MATH]\sqrt{2x-1}=[/MATH]1-[MATH]\sqrt{1-x}[/MATH] means [MATH]2x-1≤[/MATH]1 [MATH]\text{ so }\boxed{\;x≤1\;}[/MATH]
[MATH]x≥1 \text{ and } x≤1 \text{ so } x=1.[/MATH]
Now check to see if x=1 is a solution of the original equation - which it clearly is.
Subhotosh, sorry but that is simply not true. As you said, maybe with an extraneous solution. That implies, since we have a quadratic equation, that there may not be any extraneous solutions which implies that there can be two solutions. If there can be two solutions, then there will be an equation that has NO solutions.The method shown above will always produce a solution (along with - may be - extraneous solution, which you need to discard).