Hi I need help with what should be a simple algebra problem, any help would be great thanks.

the square root of 3 - the square root of 2 all over the sqaure root of (3 - the square root of 2)

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- Feb 13th 2008, 01:13 PMIlusaAlgebra surd question please help
Hi I need help with what should be a simple algebra problem, any help would be great thanks.

the square root of 3 - the square root of 2 all over the sqaure root of (3 - the square root of 2) - Feb 13th 2008, 01:42 PMPlato
$\displaystyle \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {\sqrt 3 - \sqrt 2 } }}\frac{{\sqrt {\sqrt 3 - \sqrt 2 } }}{{\sqrt {\sqrt 3 - \sqrt 2 } }} = ?

$ - Feb 14th 2008, 08:07 AMIlusa
The question is

$\displaystyle

\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {\ 3 - \sqrt 2 } }}

$

sorry I left an important bit out the answer has to be in the form

$\displaystyle a-b{{\sqrt c }}$

PS plato when you were multiplying the surd by the conjugate would the - not be outside the brackets, I'm abit confused - Feb 14th 2008, 08:16 AMtopsquark
According to the general prescription there is a minus sign. But since it cancels out, most would not include it. For example, in Plato's hint:

$\displaystyle \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {\sqrt 3 - \sqrt 2 } }} \cdot \frac{-{\sqrt {\sqrt 3 - \sqrt 2 } }}{-{\sqrt {\sqrt 3 - \sqrt 2 } }}$

$\displaystyle = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {\sqrt 3 - \sqrt 2 } }}\frac{{\sqrt {\sqrt 3 - \sqrt 2 } }}{{\sqrt {\sqrt 3 - \sqrt 2 } }}$

Now just multiply things out and see what you get.

-Dan - Feb 14th 2008, 09:07 AMIlusa
the 3 on the denominator should not have the extra square root only the 2 should have

The question is

$\displaystyle

\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {\ 3 - \sqrt 2 } }}

$

I multiplied it by $\displaystyle

\frac{{\sqrt {\ 3 - \sqrt 2 } }}{{\sqrt {\ 3 - \sqrt 2 } }}

$

and got $\displaystyle

\frac{{\sqrt {\ 3 - \sqrt 2 }.{\sqrt 3 - \sqrt 2 } }}{{ {\ 3 - \sqrt 2 } }}

$

where do you go form here I've factorised by ended up with the original equ - Feb 15th 2008, 03:17 AMIlusa
anyone please?

- Feb 15th 2008, 04:15 AMtopsquark