# Equivalent Forms

• Jul 19th 2006, 11:20 PM
zoso
Equivalent 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 AM
topsquark
Quote:

Originally Posted by zoso
$\displaystyle ^3\sqrt{2} (\sqrt{3)$ 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)

I see no way to simplify the first expression. Note: A slightly better way to code the cube root of 2 is: $\displaystyle \sqrt[3]{2}$ (Click to see the code.)

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

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

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

-Dan
• Jul 20th 2006, 05:19 AM
malaygoel
Quote:

Originally Posted by zoso
$\displaystyle ^3\sqrt{2} (\sqrt{3)$ 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)

$\displaystyle \sqrt[3]{2}\sqrt{3}=\sqrt[6]{4}\sqrt[6]{27}=\sqrt[6]{108}$

Malay
• Jul 20th 2006, 05:55 AM
Quick
Quote:

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

Here's a tutorial if you haven't seen it yet.
• Jul 20th 2006, 07:17 AM
zoso
k, I'm going to go read the tutorial, then come back and try something else out.
• Jul 20th 2006, 08:06 AM
zoso
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 AM
Quick
Quote:

Originally Posted by zoso
ok, lets try posting with fractions

Simplfy $\displaystyle \frac {1} {x + 1} - \frac {3} {x - 1}$ assuming $\displaystyle x = \pm 1$ (What is the tag for non-equivalents? that should read "assuming, x is not equal to..")

$\displaystyle x \neq \pm 1$
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 PM
zoso
Quote:

Originally Posted by zoso
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?

Anyone going to work through these two, quickly?
• Jul 20th 2006, 01:50 PM
Quick
Quote:

Originally Posted by zoso
ok, lets try posting with fractions

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

find the common denominator...
$\displaystyle \frac {1(x-1)} {(x + 1)(x-1)} - \frac {3(x+1)} {(x - 1)(x+1)}$

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 PM
zoso
So, $\displaystyle \frac {-2(x + 2)} {x^2 - 1}$ ?
• Jul 20th 2006, 02:57 PM
Quick
Quote:

Originally Posted by zoso
So, $\displaystyle \frac {-2(x + 2)} {x^2 - 1}$ ?

yes, I'm glad you realized that, it should be written as, $\displaystyle \neg2\frac {x + 2} {x^2 + 1}$ though
• Jul 21st 2006, 05:18 AM
topsquark
Quote:

Originally Posted by Quick
yes, I'm glad you realized that, it should be written as, $\displaystyle \neg2\frac {x + 2} {x^2 + 1}$ though

$\displaystyle -2 \frac {x + 2} {x^2 + 1} = - \frac {2(x + 2)} {x^2 + 1} = \frac {-2(x + 2)} {x^2 + 1}$

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 PM
Quick
Quote:

Originally Posted by topsquark
$\displaystyle -2 \frac {x + 2} {x^2 + 1} = - \frac {2(x + 2)} {x^2 + 1} = \frac {-2(x + 2)} {x^2 + 1}$

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

I think they are looking for simplest form...

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

Originally Posted by Quick
I think they are looking for simplest form...

BTW: I am starting to use the $\displaystyle \neg$ for negative numbers, particularly fractions.

But why use it? As far as I know, "-" is standard for at least complex numbers.

-Dan