## simplifying square roots

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### simplifying square roots

by ern5231 » Sun Aug 16, 2009 7:43 am
The attached questions needs to be simplified. OA later
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by mohitsharda » Sun Aug 16, 2009 8:30 am
(1+ root 3)^2 = 1+3+ (2 root 3)
= 4 + 2 root3
= 2( 2+root 3)

=> 1 + root 3 = (root 2) ( root( 2+ root 3)) ...

Substitute this value of (1 + root3) in the above expression
we get,

(root 2) ( root(2+ root3))*( root(2- root 3)) = (root 2) (root( 4-3))
{ using (a-b)(a+b) = (a^2 - b^2) }

Now, can we have the OA please :roll:
MS

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by gmatv09 » Sun Aug 16, 2009 10:39 am

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by varunkh70 » Sun Aug 16, 2009 7:28 pm
Nice thinking mohit. OA is root 2

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by winnerhere » Mon Aug 17, 2009 8:45 am
square the whole expression - thats it.

we get (after squaring)

(1+3+2âˆš3) (2-âˆš3) = 2

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by GambitOS » Thu Aug 20, 2009 12:44 am
OA is 2.

a. (1+root3)=(root1+root3)=(root4)=2
b. (root(2-root3))=(root(root4-root3)=(root1)=1

So 2*1=2

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by tohellandback » Thu Aug 20, 2009 12:54 am
GambitOS wrote:OA is 2.

a. (1+root3)=(root1+root3)=(root4)=2
b. (root(2-root3))=(root(root4-root3)=(root1)=1

So 2*1=2
so when you give your answer don't say that it's the OA until you are sure that it's the OA
now,
(root1+root3)=(root4)=2--Wrong
(root1 + root3) is not equal to root(1+3)
root(root4-root3)=(root1)=1--wrong
(root4-root3) is not equal to root(4-3)
The powers of two are bloody impolite!!

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by GambitOS » Thu Aug 20, 2009 12:57 am
Thanks! You are right, I am mistook. Sorry.

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by sacx » Thu Aug 20, 2009 2:48 am
(1+âˆš3)^1/2 * (1+âˆš3)^1/2 * [1 + (1-âˆš3)]^1/2
(1+âˆš3)^1/2 * [1+âˆš3+1-3]^1/2
(1+âˆš3)^1/2 * (âˆš3-1)^1/2
(3-1)^1/2
âˆš2
SACX

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by real2008 » Thu Aug 20, 2009 10:42 am
can't it be -sqrt(2) too?

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