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Square root of 7 + square root 48

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Senior Manager
Senior Manager
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Joined: 05 Oct 2008
Posts: 258
Square root of 7 + square root 48 [#permalink]

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New post 06 Nov 2008, 03:18
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

# 1.0
# 1.7
# 2.0
# 2.4
# 3.0

Can you please provide the full solution. Thanks

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VP
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Joined: 17 Jun 2008
Posts: 1479
Re: Square Roots [#permalink]

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New post 06 Nov 2008, 03:37
study wrote:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

# 1.0
# 1.7
# 2.0
# 2.4
# 3.0

Can you please provide the full solution. Thanks


Say \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = x\)
or, \(\sqrt{7 + \sqrt{48}} = x + \sqrt{3}\)
squaring both sides,
\(7 + 4\sqrt{3} = x^2 + 2x\sqrt{3} + 3\)
or, \(4 + 4\sqrt{3} = x(x + 2\sqrt{3})\)
or \(2(2 + 2\sqrt{3} = x(x + 2\sqrt{3})\)

Hence, x = 2.

I am sure there would be an easier way out.
Manager
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Joined: 23 Aug 2008
Posts: 62
Re: Square Roots [#permalink]

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New post 06 Nov 2008, 03:48
study wrote:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

# 1.0
# 1.7
# 2.0
# 2.4
# 3.0

Can you please provide the full solution. Thanks


I'm going for 2.0, but in an unorthodox way compared to scthakur...

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)
This is approx = \(\sqrt{14} - \sqrt{3}\) since \(\sqrt{49}=7\)
\(\sqrt{14}\) is between 3 and 4 (closer to 4), since \(\sqrt{9} < \sqrt{14} < \sqrt{16}\)

Then plugging in the answers (knowing \(\sqrt{3}\) approx = 1.7)
1.7 + 3.0 = 4.7 = too high (above 4)
1.7 + 2.4 = 4.1 = too high (above 4)
1.7 + 2.0 = 3.7 = about right
1.7 + 1.7 = 3.4 = clearly too low (since \(\sqrt{14} > \sqrt{3} , and \sqrt{3}=1.7\))

I found this method quite quick (~1 minute), although I do appreciate the algebraic method.
Manager
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Joined: 30 Sep 2008
Posts: 111
Re: Square Roots [#permalink]

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New post 06 Nov 2008, 04:26
for this type of question, usually the function under the square root is a square.

\(7+sqrt(48) = 7 + 4*sqrt(3) = 3 + 2*2*sqrt(3) + 2^2 = (sqrt(3) + 2)^2\)
Director
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Posts: 695
Re: Square Roots [#permalink]

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New post 06 Nov 2008, 05:56
lylya4 wrote:
for this type of question, usually the function under the square root is a square.

\(7+sqrt(48) = 7 + 4*sqrt(3) = 3 + 2*2*sqrt(3) + 2^2 = (sqrt(3) + 2)^2\)


Nice approach!

Though I will consider it safe to use algebraic method on the D day

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Square Roots   [#permalink] 06 Nov 2008, 05:56
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Square root of 7 + square root 48

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