$\displaystyle

^3\sqrt{2} (\sqrt{3)

$ is equal to?

an equivalent form of $\displaystyle (2\sqrt{2} - \sqrt{6})^2 $ is ?

(This is also an excuse for me to get used to the php math code, which I never used before)

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- Jul 19th 2006, 11:20 PMzosoEquivalent Forms
$\displaystyle

^3\sqrt{2} (\sqrt{3)

$ is equal to?

an equivalent form of $\displaystyle (2\sqrt{2} - \sqrt{6})^2 $ is ?

(This is also an excuse for me to get used to the php math code, which I never used before) - Jul 20th 2006, 04:45 AMtopsquarkQuote:

Originally Posted by**zoso**

$\displaystyle \left ( 2 \sqrt{2} - \sqrt{6} \right ) ^2 = \left ( 2 \sqrt{2} \right ) ^2 - 2 \cdot \left ( 2 \sqrt{2} \right ) \left ( \sqrt{6} \right ) + \left ( - \sqrt{6} \right ) ^2$

$\displaystyle = 4 \cdot 2 - 2 \cdot 2 \sqrt{12} + 6 = 14 - 4 \sqrt{12}$

$\displaystyle = 14 - 4 \cdot 2 \sqrt{3} = 14 - 8 \sqrt{3}$

-Dan - Jul 20th 2006, 05:19 AMmalaygoelQuote:

Originally Posted by**zoso**

Malay - Jul 20th 2006, 05:55 AMQuickQuote:

Originally Posted by**zoso**

- Jul 20th 2006, 07:17 AMzoso
k, I'm going to go read the tutorial, then come back and try something else out.

- Jul 20th 2006, 08:06 AMzoso
ok, lets try posting with fractions

Simplfy $\displaystyle \frac {1} {x + 1} - \frac {3} {x - 1}$ assuming $\displaystyle x \neq\pm 1$

$\displaystyle \frac {1} {j + 20} - \frac {1} {j} = 400$

$\displaystyle \frac {1} {j} - \frac {1} {j + 20} = 1$

$\displaystyle \frac {400} {j + 20} - \frac {400} {j} = 1 $

$\displaystyle \frac {400} {j} - \frac {400} {j + 20} = 1$

M drives 400 in one hour less than J. If M's speed is 20 km/h faster than J's, what is the proper equation to determine the speed of J? - Jul 20th 2006, 09:08 AMQuickQuote:

Originally Posted by**zoso**

or

$\displaystyle x \not =\pm 1$

something interesting:

$\displaystyle \frac{8}{4}\quad\Rightarrow\quad\frac{\not2\times \not2\times 2}{\not2\times \not2}\quad\Rightarrow\quad2$ - Jul 20th 2006, 01:18 PMzosoQuote:

Originally Posted by**zoso**

- Jul 20th 2006, 01:50 PMQuickQuote:

Originally Posted by**zoso**

$\displaystyle \frac {1(x-1)} {(x + 1)(x-1)} - \frac {3(x+1)} {(x - 1)(x+1)}$

multiply: $\displaystyle \frac {x-1} {x^2 + 1} - \frac {3x+3} {x^2 + 1}$

subtract: $\displaystyle \frac {x-1-3x-3} {x^2 + 1}$

group like terms: $\displaystyle \frac {x-3x-1-3} {x^2 + 1}$

subtract: $\displaystyle \frac {\neg 2x-4} {x^2 + 1}$ - Jul 20th 2006, 02:37 PMzoso
So, $\displaystyle \frac {-2(x + 2)} {x^2 - 1}$ ?

- Jul 20th 2006, 02:57 PMQuickQuote:

Originally Posted by**zoso**

- Jul 21st 2006, 05:18 AMtopsquarkQuote:

Originally Posted by**Quick**

I don't see why any one form should be more "correct" than any other. As far as I know there is no consensus on any particular form.

Also, Quick: Why do you keep using "$\displaystyle \neg$" as opposed to simply "-"?

-Dan - Jul 21st 2006, 06:36 PMQuickQuote:

Originally Posted by**topsquark**

BTW: I am starting to use the $\displaystyle \neg$ for negative numbers, particularly fractions. - Jul 25th 2006, 03:46 AMtopsquarkQuote:

Originally Posted by**Quick**

-Dan