1. May 15, 2012

AnTiFreeze3

1. The problem statement, all variables and given/known data
The expression can be simplified. Simplify the following....

2. Relevant equations
1
_____________ - Starting equation

√3 + √2 - √5

3. The attempt at a solution

1 * [(√3 + √2) + √5]
_____________ - Multiplied by conjugate

√3 + √2 - √5 * [(√3 + √2) + √5]

√3 + √2 + √5
________________ - Next I foiled (√3 + √2)^2 and ended up with:

(√3 + √2)^2 + √5

√3 + √2 + √5
______________ - This is where I got stuck......

11 + 6√2 + √5

I tried canceling out the √5 (I honestly have no clue whether you're allowed to do that at this point in the equation, please explain to me why or why not you would be able to do that) and when I did, my final solution was nowhere near the answer in the book.

I'm usually fine for problems where I need to turn a trinomial into a binomial using the associative property in order to use the conjugate to simplify, but this is the first problem that we have had where all three terms on the bottom are radicals. Did I maybe make a mistake when foiling (√3 + √2)^2... (foiling, or FOIL, is a term we use at my school, meaning First, Inner, Outer, Last, I have no idea if that's a widespread term for how to solve binomials being multiplied by each other), or am I just not seeing something else that I need to do?

By the way, the answer that my book had in the back was:

(3√2 + 2√3 + √30)
_________________

12

2. May 15, 2012

scurty

Okay, good so far.. (you might want to watch your parentheses in your last line though)

Okay, that step is wrong.

$\begin{eqnarray*} [(\sqrt{3} + \sqrt{2}) + \sqrt{5}] \cdot [(\sqrt{3} + \sqrt{2}) - \sqrt{5}] &=& (\sqrt{3} + \sqrt{2})^2 - [(\sqrt{3} + \sqrt{2}) \cdot \sqrt{5}] + [(\sqrt{3} + \sqrt{2}) \cdot \sqrt{5}] - \sqrt{5}^2 \\ &=& (\sqrt{3} + \sqrt{2})^2 - 5 \end{eqnarray*}$

Seemed like a simple algebra error. Square the term in the denominator and the rest of the problem should be pretty easy!

3. May 15, 2012

AnTiFreeze3

Alright thanks, I guess I forgot to multiply the √5 by √5.

4. May 15, 2012