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# 20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =

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Math Expert
Joined: 02 Sep 2009
Posts: 54371
20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =  [#permalink]

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10 Sep 2017, 05:20
00:00

Difficulty:

5% (low)

Question Stats:

89% (00:58) correct 11% (01:36) wrong based on 91 sessions

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$$\sqrt{20} + \sqrt{45} - \sqrt{5} - \sqrt{80}=$$

A. $$-2\sqrt{5}$$

B. $$-\sqrt{10}$$

C. 1

D. $$\sqrt{5}$$

E. 0

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Joined: 22 May 2016
Posts: 2639
20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =  [#permalink]

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10 Sep 2017, 07:55
Bunuel wrote:
$$\sqrt{20} + \sqrt{45} - \sqrt{5} - \sqrt{80}=$$

A. $$-2\sqrt{5}$$

B. $$-\sqrt{10}$$

C. 1

D. $$\sqrt{5}$$

E. 0

$$\sqrt{20} + \sqrt{45} - \sqrt{5} - \sqrt{80}=$$

Use the rule that $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$

Factor values under the radical signs

$$\sqrt{4*5} + \sqrt{9*5} - \sqrt{5} - \sqrt{16*5}=$$

$$2\sqrt{5} + 3\sqrt{5} - \sqrt{5} - 4\sqrt{5} = 0$$

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Re: 20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =  [#permalink]

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10 Sep 2017, 08:43
Bunuel wrote:
$$\sqrt{20} + \sqrt{45} - \sqrt{5} - \sqrt{80}=$$

A. $$-2\sqrt{5}$$

B. $$-\sqrt{10}$$

C. 1

D. $$\sqrt{5}$$

E. 0

$$\sqrt{20} + \sqrt{45} - \sqrt{5} - \sqrt{80}$$

$$\sqrt{2*2*5} + \sqrt{3*3*5} - \sqrt{5} - \sqrt{4*4*5}$$

$$2\sqrt{5} + 3\sqrt{5} - \sqrt{5} - 4\sqrt{5}$$

$$\sqrt{5} (2 + 3 - 1 - 4)$$

$$\sqrt{5} (5-5)$$

$$\sqrt{5} (0) = 0$$

Re: 20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =   [#permalink] 10 Sep 2017, 08:43
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# 20^(1/2) + 45^(1/2) - 5^(1/2) - 80^(1/2) =

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