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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

I'm having trouble doing this problem:

solve for x

(square root of ( x + 1) + (square root of (x - 1) 4x - 1

_________(divided by)_____________________ = ___(divided by)___

(square root of ( x + 1) - (square root of (x - 1) (2)

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

I will work on it ...

I think the trick will be to multiply by [sqrt(x+1)-sqrt(x-1)]^2

Anyway, let's try and see how far we get

[sqrt(x+1)-sqrt(x-1)] * [sqrt(x+1)+sqrt(x-1)] = [sqrt(x+1)-sqrt(x-1)]^2 * (4x-1)/2

sqrt(x+1) * sqrt(x+1) + sqrt(x+1) * sqrt(x-1) - sqrt(x-1) * sqrt(x-1) - sqrt(x-1) * sqrt(x-1) = ...

(x+1) + (terms cancel each other) - (x-1) = ...

2 = [sqrt(x+1)-sqrt(x-1)] * [sqrt(x+1)-sqrt(x-1)] * (4x-1)/2

2 = [sqrt(x+1) * sqrt(x+1) - sqrt(x+1) * sqrt(x-1) - sqrt(x-1) * sqrt(x+1) + sqrt(x-1) * sqrt(x-1)] * (4x-1)/2

2 = [(x+1) - 2 * sqrt(x+1) * sqrt(x-1) + (x-1)] * (4x-1)/2

2 = [2x - 2 * sqrt(x+1) * sqrt(x-1) ] * (4x-1) / 2

4 = [2x - 2 * sqrt(x+1) * sqrt(x-1) ] * (4x-1)

Hmmm ... that helped a bit, let's try it again but using [sqrt(x+1)+sqrt(x-1)] * [sqrt(x+1)-sqrt(x-1)]

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

For simplicity, let's call sqrt(x+1) "A" and sqrt(x-1) "B"

You have: (A+B)/(A-B) = (4x-1)/2

Multiply both sides by (A+B)*(A-B): (A+B)*(A-B) * (A+B)/(A-B) = (A+B)*(A-B) * (4x-1)/2

Cancel: (A+B)^2 = (A+B)*(A-B) * (4x-1)/2

Expand: A^2 + 2AB + B^2 = (A^2 - B^2) * (4x-1)/2

Substitute A^2 = (x+1) and B^2 = (x-1): (x+1) + 2AB + (x-1) = ((x+1) - (x-1)) * (4x-1)/2

Simplify: 2x + 2AB = 2 * (4x-1)/2

More: 2x + 2AB = 4x - 1

More: 2AB = 2x - 1

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

Wow it makes alot more sense putting it that way, I was tying to aproach it from some other way, which turned out horrible.Thanks for the help I understand it now

*Last edited by im really bored (2005-05-12 12:42:09)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Well, we still aren't ALL the way there ... we still have 2AB to solve

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

I know, but i can handle that. It was just my approach that was getting me. I was trying to go at it in a more complicated way

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Cool ... I will leave it for someone else then ...

Good luck.

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

2AB = 2x - 1 ^2

4A^2B^2=4x^2 - 2x + 1

Now: A=sqr(x+1); B=sqr(x-1)

4x^2-4=4x^2 - 4x + 1

then:

4x=5

x=5/4

It was easy, because administrator has finished the difficult part.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Thanks, Milos ...

TEAMWORK !

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

n e time

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