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Intern
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Joined: 05 Apr 2012
Posts: 47

Kudos [?]: 35 [0], given: 12

SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 05:42
Hello
can anyone help me to work with the following exercises

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


Best Regards

Kudos [?]: 35 [0], given: 12

Director
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Re: SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 06:52
keiraria wrote:
Hello
can anyone help me to work with the following exercises

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


Best Regards


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

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

Kudos [?]: 487 [0], given: 44

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

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Re: SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 14:11
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
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Re: SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 14:46
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 [?]: 35 [0], given: 12

Intern
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Joined: 05 Apr 2012
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Re: SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 14:47
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 [?]: 35 [0], given: 12

Director
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Re: SQUARE ROOTS [#permalink]

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New post 11 Apr 2017, 18:09
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?
_________________

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

Kudos [?]: 487 [0], given: 44

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Re: SQUARE ROOTS [#permalink]

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New post 12 Apr 2017, 12:04
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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 [?]: 208 [1] , given: 46

Intern
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Kudos [?]: 35 [0], given: 12

Re: SQUARE ROOTS [#permalink]

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New post 14 Apr 2017, 01:05
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
IMG_3709.JPG [ 1.62 MiB | Viewed 308 times ]

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Re: SQUARE ROOTS [#permalink]

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New post 14 Apr 2017, 06:41

Kudos [?]: 487 [0], given: 44

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Re: SQUARE ROOTS [#permalink]

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New post 19 Apr 2017, 14:00
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
IMG_3711 (2).JPG [ 1.52 MiB | Viewed 242 times ]

Kudos [?]: 35 [0], given: 12

Director
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Status: Long way to go!
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Kudos [?]: 487 [0], given: 44

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Re: SQUARE ROOTS [#permalink]

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New post 20 Apr 2017, 01:16
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.

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

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

Kudos [?]: 487 [0], given: 44

Re: SQUARE ROOTS   [#permalink] 20 Apr 2017, 01:16
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