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keiraria
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SQUARE ROOTS 11 Apr 2017, 05:42
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Hello
can anyone help me to work with the following exercises

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


Best Regards

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SQUARE ROOTS 11 Apr 2017, 06:52
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keiraria
Hello
can anyone help me to work with the following exercises

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


Best Regards
Show more

You mean \((\sqrt{6}+2)(\sqrt{3} - 2)\) ?
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laxpro2001
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SQUARE ROOTS 11 Apr 2017, 14:11
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nguyendinhtuong
keiraria
Hello
can anyone help me to work with the following exercises

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


Best Regards

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


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?
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SQUARE ROOTS 11 Apr 2017, 14:46
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nguyendinhtuong
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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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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 ?
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SQUARE ROOTS 11 Apr 2017, 14:47
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nguyendinhtuong
keiraria
Hello
can anyone help me to work with the following exercises

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


Best Regards

You mean \((\sqrt{6}+2)(\sqrt{3} - 2)\) ?
Show more
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 ?
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SQUARE ROOTS 11 Apr 2017, 18:09
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keiraria
nguyendinhtuong
keiraria
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 ?
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Sorry, but I can't get your point. Can you elaborate more?
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SQUARE ROOTS 12 Apr 2017, 12:04
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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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SQUARE ROOTS 14 Apr 2017, 01:05
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ccooley
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.
Show more
ccooley
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.
Show more

hello
thanks for your answer and feedback, from the beginning this simple exercise has something wrong
I have included my approach as attachment
thanks
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File comment: my approach
IMG_3709.JPG
IMG_3709.JPG [ 1.62 MiB | Viewed 2966 times ]

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broall
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SQUARE ROOTS 14 Apr 2017, 06:41
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OMG, could you post another better image? I cant read anything
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keiraria
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SQUARE ROOTS 19 Apr 2017, 14:00
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nguyendinhtuong
OMG, could you post another better image? I cant read anything
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HELLO
PLEASE FIND ATTACHED A BETTER IMAGE

THANKS

BEST REGARDS
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File comment: MY APPROACH
IMG_3711 (2).JPG
IMG_3711 (2).JPG [ 1.52 MiB | Viewed 2854 times ]

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broall
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SQUARE ROOTS 20 Apr 2017, 01:16
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keiraria

HELLO
PLEASE FIND ATTACHED A BETTER IMAGE

THANKS

BEST REGARDS
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The OA is totally wrong.

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

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Hi there,
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