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

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Math Expert
Joined: 02 Sep 2009
Posts: 52108
12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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06 Apr 2018, 04:09
00:00

Difficulty:

15% (low)

Question Stats:

84% (01:04) correct 16% (02:18) wrong based on 77 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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Joined: 22 May 2016
Posts: 2326
12^(1/2) + 108^(1/2) + 48^(1/2) =  [#permalink]

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06 Apr 2018, 20:20
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) = &nbs [#permalink] 06 Apr 2018, 20:20
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