Is it 6?

I think that the main thing here, as in most algebraic stuff that have fractions is to recognize the a^2-b^2 = (a-b)(a+b). Then, we can multiply the nominator and denominator by 1, or by the (a-b) and we don't change the meaning of the equation.

2rt3+3rt2 rt2-rt3 2rt3rt2-2*3+3*2-3rt2rt3
--------- * ------- = -----------------------
rt2+rt3 rt2-rt3 2-3

-1*rt2rt3
--------- = rt2rt3 However, it was all squared, so it
-1

is (rt2rt3)^2=2*3=6

I'm also new to this, and unsure about the answer, but hope it's ok.

Leon1984 wrote:Is it 6?

I think that the main thing here, as in most algebraic stuff that have fractions is to recognize the a^2-b^2 = (a-b)(a+b). Then, we can multiply the nominator and denominator by 1, or by the (a-b) and we don't change the meaning of the equation.

2rt3+3rt2 rt2-rt3 2rt3rt2-2*3+3*2-3rt2rt3
--------- * ------- = -----------------------
rt2+rt3 rt2-rt3 2-3

-1*rt2rt3
--------- = rt2rt3 However, it was all squared, so it
-1

is (rt2rt3)^2=2*3=6

I'm also new to this, and unsure about the answer, but hope it's ok.

Hi Leon, I tried hardly to follow your steps I but can't get it.

Can you make it more clear ? and add parentheses pls ..

Sure, I'll just do it in few steps:
When we multiply any number, and a fraction, by 1, it value does not change. Note, that 1 can be a fraction, so we can multiply the nominator and denominator by the (a-b) and we don't change the meaning of the equation.

2rt3+3rt2 rt2-rt3
--------- * ------- = multiplying by 1 to get a^2-b^2
rt2+rt3 rt2-rt3

(2rt3+3rt2)*(rt2-rt3)
---------------------
(rt2+rt3)*(rt2-(rt3)

2rt3rt2-2*3+3*2-3rt2rt3
-----------------------
(rt2)^2 - (rt3)^2

2*(rt3rt2)-3*(rt2rt3)
---------------------
2-3

-rt2rt3
-------
-1

(rt2rt3)^2 We square it since, if you recall, the whole expression in the beginning was squared, and so far we've been working with the internal part of it.

(rt2*rt3)(rt2*rt3)=2*3=6

I hope I've made it more clear.

GMATER911 wrote:( ( 2√3+3√2)/√2 + √3) ^2 = ??


Unfortunately, I don't have the answer. Can someone solve it.

The main idea is to recognize (A+B)^2 = A^2 + 2AB + B^2 ... but I can't get an integer .. what"s wrong?

parenthesis are wrong :

( ( 2√3+3√2)/√2 + √3) ^2 = ??

should be :

( ( 2√3+3√2)/(√2 + √3) ) ^2 = ??

mridul_dave wrote:

GMATER911 wrote:( ( 2√3+3√2)/√2 + √3) ^2 = ??


Unfortunately, I don't have the answer. Can someone solve it.

The main idea is to recognize (A+B)^2 = A^2 + 2AB + B^2 ... but I can't get an integer .. what"s wrong?

parenthesis are wrong :

( ( 2√3+3√2)/√2 + √3) ^2 = ??

should be :

( ( 2√3+3√2)/(√2 + √3) ) ^2 = ??

Hi:
I changed to:

( (2√3+3√2 + √2.√3)/√2 )^2

that is, the √2 as common denominator for √3 too, and I got 18, am I wrong?

Silvia

Actually if you solve on the basis of (a + b)^2 its easy, answer should be 6

second last equation I got is
30 + 12*sq.root(6)/(5 + 2 * sq.rt(6))

from above take 6 common so that will be
6 * (5 + 2 * sq.rt(6)) / (5 + 2 * sq.rt(6))

so answer = 6

Here another easy way to solve..

{(2 sqrt3 + 3 sqrt2)/ (sqrt2 + sqrt3) }^2 = ?

1) Factor the nomerator to get the same denominator's expression.

{ (sqrt2*sqrt3)(sqrt2+sqrt3) /(sqrt2+sqrt3)}^2
to double check you factorization multiply the bolded expression, so you should get (2 sqrt3 + 3 sqrt2 )

2) Cancel out the top and bottom (sqrt2+sqrt3)

=(sqrt2 * sqrt3)^2
= 2*3
= 6 ... this is the answer.