$\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.)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
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?
$\displaystyle x \neq \pm 1$Originally Posted by zoso
or
$\displaystyle x \not =\pm 1$
something interesting:
$\displaystyle \frac{8}{4}\quad\Rightarrow\quad\frac{\not2\times \not2\times 2}{\not2\times \not2}\quad\Rightarrow\quad2$
find the common denominator...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: $\displaystyle \frac {\neg 2x-4} {x^2 + 1}$
$\displaystyle -2 \frac {x + 2} {x^2 + 1} = - \frac {2(x + 2)} {x^2 + 1} = \frac {-2(x + 2)} {x^2 + 1}$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