Stuck...

$\displaystyle 25\sqrt {3 } + 3\sqrt {3} - \sqrt {3}$
-----------------
$\displaystyle 2 + 2 - 2$

$\displaystyle 28\sqrt {3}$
over 2

2. Originally Posted by fluffy_penguin
Stuck...

$\displaystyle 25\sqrt {3 } + 3\sqrt {3} - \sqrt {3}$
-----------------
$\displaystyle 2 + 2 - 2$

$\displaystyle 28\sqrt {3}$
over 2
Well... Your math was slightly flawed.

$\displaystyle (5\sqrt {3 } + 3\sqrt {3} - \sqrt {3})/2$

Note the denominator is just the common divisor... not a sum/different/etc.

Also note: $\displaystyle \sqrt {75}$ equals $\displaystyle \sqrt {5^2*3}$ equals $\displaystyle 5\sqrt {3}$
This means you can factor out a 5 (the root of 25) and not 25.

$\displaystyle (5\sqrt {3 } + 3\sqrt {3} - \sqrt {3})/2$
equals
$\displaystyle ((5+3-1)\sqrt {3})/2$
$\displaystyle (7\sqrt {3})/2$

3. $\displaystyle \begin{array}{rcl} \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }} & = & \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }}\left( {\frac{{\sqrt 4 }}{{\sqrt 4 }}} \right) \\ & = & \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt {12} }}{4} \\ \end{array}$

$\displaystyle \begin{array}{rcl} \frac{{\sqrt {75} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{{\sqrt 4 }}& = & \frac{{\sqrt {3 \cdot 25} }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{\sqrt {4 \cdot 3} }}{4} \\ & = & \frac{{5\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{2} - \frac{{2\sqrt 3 }}{4} \\ \end{array}$

4. Are you only taking positive roots? Because $\displaystyle \frac{9\sqrt{3}}{2}$ is also
a solution.

5. Originally Posted by SnipedYou
Are you only taking positive roots? Because $\displaystyle \frac{9\sqrt{3}}{2}$ is also
a solution.
What in the world can that statement mean?

6. Originally Posted by SnipedYou
Are you only taking positive roots? Because $\displaystyle \frac{9\sqrt{3}}{2}$ is also
a solution.
Roots? There are no functions here. And there is only one answer for this problem.

At a guess, perhaps you are confused about taking $\displaystyle \pm$ when you take a square root? When you are looking at a numeric (that is to say, non-variable) expression involving a radical like a square root, you take the positive sign. This is a standard definition.

-Dan

7. Originally Posted by topsquark
At a guess, perhaps you are confused about taking $\displaystyle \pm$ when you take a square root?
Dan, what a nice guy you must be. That was such a polite way to put it.
I wanted to say “mathematical amateurs ought to stay out of it.”

8. Originally Posted by Plato
Dan, what a nice guy you must be. That was such a polite way to put it.
I wanted to say “mathematical amateurs ought to stay out of it.”

-Dan