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Math Expert V
Joined: 02 Sep 2009
Posts: 59075
12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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Difficulty:   5% (low)

Question Stats: 92% (01:14) correct 8% (02:10) wrong based on 130 sessions

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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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Senior SC Moderator V
Joined: 22 May 2016
Posts: 3657
12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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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:

$$2\sqrt{3} + 6\sqrt{3} + 4\sqrt{3} =12\sqrt{3}$$

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12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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1
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.

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Joined: 23 Apr 2018
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12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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1
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]

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1
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]

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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

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: 12^(1/2) + 108^(1/2) + 48^(1/2) =   [#permalink] 01 Jul 2019, 17:38
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