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12^(1/2) + 108^(1/2) + 48^(1/2) =

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Bunuel
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12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   06 Apr 2018, 05:09
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\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)


A. \(12\sqrt{3}\)

B. 24

C. \(12\sqrt{5}\)

D. \(48\sqrt{3}\)

E. \(\sqrt{168}\)
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12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   06 Apr 2018, 21:20
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Bunuel wrote:
\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)

A. \(12\sqrt{3}\)

B. 24

C. \(12\sqrt{5}\)

D. \(48\sqrt{3}\)

E. \(\sqrt{168}\)

\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)

Put factors under the radical signs, looking for factors that are perfect squares.
Look also, and first, for a common factor that is NOT a perfect square, one that will remain beneath the radical sign.

Check the answer choices for possible factors that are not perfect squares.
Here, choices are \(\sqrt{3}\), \(\sqrt{5}\), and \(\sqrt{168}\)

5 is not a factor at all. 168 is too great to work with. Ignore.
Try 3 under the radical sign.
Once you divide each number by 3, the remaining factor is a perfect square.

\(\sqrt{3*4} + \sqrt{3*36} + \sqrt{3*16} =\)

\((\sqrt{3}*\sqrt{4}) + (\sqrt{3}*\sqrt{36}) + (\sqrt{3}*\sqrt{16})\)

Take the square roots of the factors that are perfect squares:

\(2\sqrt{3} + 6\sqrt{3} + 4\sqrt{3} =12\sqrt{3}\)

Answer A
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12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   27 Jun 2019, 09:39
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Convert every term in Root (12).
First term: Root 12
2nd term: Root 48= Root(4*12)=2 Root 12

3rd term: Root 109=Root (9*12)=3 Root 12.

So, 1st term + 2nd term +3rd term
=6 Root 12 (since no such thing available in options, do further steps.
= 6 * Root (4*3)
=6*2*Root 3
=12 ROOT 3.
A is my answer.

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12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   Updated on: 27 Jun 2019, 11:36
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Bunuel wrote:
\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)


A. \(12\sqrt{3}\)

B. 24

C. \(12\sqrt{5}\)

D. \(48\sqrt{3}\)

E. \(\sqrt{168}\)



The answer options are varied, which is a hint, once we find the common factor, we can quickly guess and move on

solving the biggest value first so we'll know the maximum value out of the 3 terms.
\(\sqrt{108}\) =
\(\sqrt{36*3}\)
\(6\sqrt{3}\),

as we can see A is the smaller of the two answers with \(\sqrt{3}\).. we can mark A right away..

there's no chance that we will touch 48 outside square root, as biggest number outside square root is 6

also you can guess that \(\sqrt{12}\) will be somewhat closer to 3
and \(\sqrt{48}\) will be closer to 7
and as we have \(\sqrt{3}\) as common, so these 2 values will be les than 3 and 7
6+7+3 = 16, closest one is \(12\sqrt{3}\)
so A is pretty much right on this one

Originally posted by Shrey9 on 27 Jun 2019, 11:07.
Last edited by Shrey9 on 27 Jun 2019, 11:36, edited 1 time in total.
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Re: 12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   27 Jun 2019, 11:27
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Bunuel wrote:
\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)


A. \(12\sqrt{3}\)

B. 24

C. \(12\sqrt{5}\)

D. \(48\sqrt{3}\)

E. \(\sqrt{168}\)


\(\sqrt{12} + \sqrt{108} + \sqrt{48}\)
--> \(\sqrt{4*3} + \sqrt{36*3} + \sqrt{16*3}\)
--> \(2\sqrt{3} + 6\sqrt{3} + 4\sqrt{3}\)
--> \(12\sqrt{3}\)

IMO Option A

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Re: 12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   01 Jul 2019, 17:38
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Bunuel wrote:
\(\sqrt{12} + \sqrt{108} + \sqrt{48} =\)


A. \(12\sqrt{3}\)

B. 24

C. \(12\sqrt{5}\)

D. \(48\sqrt{3}\)

E. \(\sqrt{168}\)


Simplifying and combining, we have:

√4 x √3 + √36 x √3 + √16 x √3

2√3 + 6√3 + 4√3 = 12√3

Answer: A
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Re: 12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink] New post   14 Mar 2024, 09:42
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Re: 12^(1/2) + 108^(1/2) + 48^(1/2) = [#permalink]
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