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)
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
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
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