Can't solve this equation ,pls help
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Dr.Peterson
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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).
That's a big reason we ask you to show work:
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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.
[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.
Last edited:
D
Deleted member 4993
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Following Jomo's advice:
√(2x-1) = 1 - √(x-1) .............(1)................ square both sides
(2x-1) = 1 + (x-1) - 2*√(x-1) ................. simplify
2√(x-1) = 1 - x ........................................... square both sides
4(x-1) = 1 + x^2 - 2x .............................simplify
x^2 - 6x + 5 = 0 ......................................factorize
(x - 5)(x - 1) = 0 ....................................... two solutions: x = 5 and x = 1
Assuming √ is always positive → x\(\displaystyle \ \ne \\\)5 ............... here x=5 is an extraneous solution (created due to 'squaring')
The other solution: x = 1 satisfies the given equation (#1).
Hence the solution is x = 1.
The method shown above will always produce a solution, if solution/s exist (along with - may be - extraneous solution, which you need to discard). Although Lex's method in response #4 is very clever and concise, it is difficult to 'see' for some problems. ......................edited
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Steven G
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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.
Always be careful when you say will always.
This error really deserves some corner time to think about this!
I suppose it is good to practise thinking skills by trying to look at things differently. (It's also fun)!
[MATH]\sqrt{1-2x}=[/MATH]1-[MATH]\sqrt{1+x}[/MATH]
so [MATH]1-2x≤1[/MATH] i.e.[MATH] \boxed{\;x≥0\;}[/MATH]
[MATH]1-\sqrt{1+x}=\sqrt{1-2x}≥0[/MATH]
so [MATH]\sqrt{1+x}≤1[/MATH], so [MATH]\boxed{\;x≤0\;}[/MATH]
[MATH]x=0[/MATH] is the only possible solution, but clearly is not a solution of the original equation.
Therefore, no solution.
[MATH]\sqrt{1-2x}=[/MATH]1-[MATH]\sqrt{1+x}[/MATH]
so [MATH]1-2x≤1[/MATH] i.e.[MATH] \boxed{\;x≥0\;}[/MATH]
[MATH]1-\sqrt{1+x}=\sqrt{1-2x}≥0[/MATH]
so [MATH]\sqrt{1+x}≤1[/MATH], so [MATH]\boxed{\;x≤0\;}[/MATH]
[MATH]x=0[/MATH] is the only possible solution, but clearly is not a solution of the original equation.
Therefore, no solution.