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.
Arithmetic Problem
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Abdulla
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Hi Leon, I tried hardly to follow your steps I but can't get it.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.
Can you make it more clear ? and add parentheses pls ..
Abdulla
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Leon1984
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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.
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.
Leon
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mridul_dave
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parenthesis are wrong :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?
( ( 2√3+3√2)/√2 + √3) ^2 = ??
should be :
( ( 2√3+3√2)/(√2 + √3) ) ^2 = ??
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ssuarezo
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mridul_dave wrote:parenthesis are wrong :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?
( ( 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
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satish.nagdev
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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
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
Wounded by GMAT but not dead
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Abdulla
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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.
{(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.
Abdulla
