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# SQUARE ROOTS

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Intern
Joined: 05 Apr 2012
Posts: 45

Kudos [?]: 37 [0], given: 12

SQUARE ROOTS [#permalink]

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11 Apr 2017, 04:42
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

Kudos [?]: 37 [0], given: 12

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1294

Kudos [?]: 1103 [0], given: 62

Location: Viet Nam
Re: SQUARE ROOTS [#permalink]

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11 Apr 2017, 05:52
keiraria wrote:
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

You mean $$(\sqrt{6}+2)(\sqrt{3} - 2)$$ ?
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Kudos [?]: 1103 [0], given: 62

Intern
Joined: 30 Mar 2017
Posts: 39

Kudos [?]: 12 [0], given: 47

Location: United States (FL)
Re: SQUARE ROOTS [#permalink]

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11 Apr 2017, 13:11
nguyendinhtuong wrote:
keiraria wrote:
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

You mean $$(\sqrt{6}+2)(\sqrt{3} - 2)$$ ?

I assume you meant to put what nguyendinhtuong wrote in his reply. In this case, you would FOIL the 2 expressions:
$$\sqrt{18} - 2\sqrt{6} + 2\sqrt{3} - 4$$
$$\sqrt{18}$$ can be re-written as $$\sqrt{9}*\sqrt{2}$$which reduces to $$3\sqrt{2}$$

Is this what you were looking for?

Kudos [?]: 12 [0], given: 47

Intern
Joined: 05 Apr 2012
Posts: 45

Kudos [?]: 37 [0], given: 12

Re: SQUARE ROOTS [#permalink]

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11 Apr 2017, 13:46
nguyendinhtuong wrote:
keiraria wrote:
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

You mean $$(\sqrt{6}+2)(\sqrt{3} - 2)$$ ?

YES THANKS AND HOW TO PROCEED THE REST THE OA IS SQUARE ROOT OF 2

I Have done 3 square root of 2 then I have put -2 square root of 3 time square root of 2 and +2 square root of 3 and- 4 ?

Kudos [?]: 37 [0], given: 12

Intern
Joined: 05 Apr 2012
Posts: 45

Kudos [?]: 37 [0], given: 12

Re: SQUARE ROOTS [#permalink]

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11 Apr 2017, 13:47
nguyendinhtuong wrote:
keiraria wrote:
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

You mean $$(\sqrt{6}+2)(\sqrt{3} - 2)$$ ?

YES THANKS AND HOW TO PROCEED THE REST THE OA IS SQUARE ROOT OF 2

I Have done 3 square root of 2 then I have put -2 square root of 3 time square root of 2 and +2 square root of 3 and- 4 ?

Kudos [?]: 37 [0], given: 12

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1294

Kudos [?]: 1103 [0], given: 62

Location: Viet Nam
Re: SQUARE ROOTS [#permalink]

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11 Apr 2017, 17:09
keiraria wrote:
nguyendinhtuong wrote:
keiraria wrote:
Hello
can anyone help me to work with the following exercises

( \sqrt{}6+2) (\sqrt{}3-2)
THANKS

Best Regards

You mean $$(\sqrt{6}+2)(\sqrt{3} - 2)$$ ?

YES THANKS AND HOW TO PROCEED THE REST THE OA IS SQUARE ROOT OF 2

I Have done 3 square root of 2 then I have put -2 square root of 3 time square root of 2 and +2 square root of 3 and- 4 ?

Sorry, but I can't get your point. Can you elaborate more?
_________________

Kudos [?]: 1103 [0], given: 62

Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 432

Kudos [?]: 277 [1], given: 64

GMAT 1: 790 Q51 V49
GRE 1: 340 Q170 V170
Re: SQUARE ROOTS [#permalink]

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12 Apr 2017, 11:04
1
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Expert's post
I think there's a mistake either with the OA (sqrt(2)), or with the phrasing of the question. As it stands, the answer to this one can't be sqrt(2). Among other issues, the answer to your question will be a negative number, since sqrt(3) < 2. So, sqrt(3) - 2 will be negative. Since it's being multiplied by a positive value, the result will be negative.
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Kudos [?]: 277 [1], given: 64

Intern
Joined: 05 Apr 2012
Posts: 45

Kudos [?]: 37 [0], given: 12

Re: SQUARE ROOTS [#permalink]

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14 Apr 2017, 00:05
ccooley wrote:
I think there's a mistake either with the OA (sqrt(2)), or with the phrasing of the question. As it stands, the answer to this one can't be sqrt(2). Among other issues, the answer to your question will be a negative number, since sqrt(3) < 2. So, sqrt(3) - 2 will be negative. Since it's being multiplied by a positive value, the result will be negative.
ccooley wrote:
I think there's a mistake either with the OA (sqrt(2)), or with the phrasing of the question. As it stands, the answer to this one can't be sqrt(2). Among other issues, the answer to your question will be a negative number, since sqrt(3) < 2. So, sqrt(3) - 2 will be negative. Since it's being multiplied by a positive value, the result will be negative.

hello
thanks for your answer and feedback, from the beginning this simple exercise has something wrong
I have included my approach as attachment
thanks
Attachments

File comment: my approach

IMG_3709.JPG [ 1.62 MiB | Viewed 449 times ]

Kudos [?]: 37 [0], given: 12

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1294

Kudos [?]: 1103 [0], given: 62

Location: Viet Nam
Re: SQUARE ROOTS [#permalink]

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14 Apr 2017, 05:41
OMG, could you post another better image? I cant read anything
_________________

Kudos [?]: 1103 [0], given: 62

Intern
Joined: 05 Apr 2012
Posts: 45

Kudos [?]: 37 [0], given: 12

Re: SQUARE ROOTS [#permalink]

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19 Apr 2017, 13:00
nguyendinhtuong wrote:
OMG, could you post another better image? I cant read anything

HELLO
PLEASE FIND ATTACHED A BETTER IMAGE

THANKS

BEST REGARDS
Attachments

File comment: MY APPROACH

IMG_3711 (2).JPG [ 1.52 MiB | Viewed 383 times ]

Kudos [?]: 37 [0], given: 12

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1294

Kudos [?]: 1103 [0], given: 62

Location: Viet Nam
Re: SQUARE ROOTS [#permalink]

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20 Apr 2017, 00:16
keiraria wrote:
nguyendinhtuong wrote:
OMG, could you post another better image? I cant read anything

HELLO
PLEASE FIND ATTACHED A BETTER IMAGE

THANKS

BEST REGARDS

The OA is totally wrong.

$$\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4 \\ = \sqrt{9 \times 2} + 2\sqrt{3} - 2\sqrt{3} \times \sqrt{2} - 4 \\ \neq \sqrt{9 \times 2} + (2\sqrt{3} - 2\sqrt{3}) - 4 \sqrt{2}$$
_________________

Kudos [?]: 1103 [0], given: 62

Re: SQUARE ROOTS   [#permalink] 20 Apr 2017, 00:16
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# SQUARE ROOTS

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