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# If a^(1/2) = 3 and b^(1/2) = 4, then (a^(1/2) + b^(1/2)/(a + b)^(1/2)

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Joined: 26 Dec 2015
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Location: United States (CA)
Concentration: Finance, Strategy
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If a^(1/2) = 3 and b^(1/2) = 4, then (a^(1/2) + b^(1/2)/(a + b)^(1/2)  [#permalink]

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Updated on: 13 Mar 2018, 21:18
2
1
00:00

Difficulty:

25% (medium)

Question Stats:

83% (01:07) correct 17% (01:44) wrong based on 47 sessions

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If $$\sqrt{a} = 3$$ and $$\sqrt{b} = 4$$, then $$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a + b}} =$$

A. 1

B. 8/7

C. 7/5

D. 12/5

E. 5

Attachment:

tpr.png [ 9.6 KiB | Viewed 798 times ]

Originally posted by LakerFan24 on 13 Mar 2018, 20:09.
Last edited by Bunuel on 13 Mar 2018, 21:18, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 58402
If a^(1/2) = 3 and b^(1/2) = 4, then (a^(1/2) + b^(1/2)/(a + b)^(1/2)  [#permalink]

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13 Mar 2018, 21:25
1
LakerFan24 wrote:
If $$\sqrt{a} = 3$$ and $$\sqrt{b} = 4$$, then $$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a + b}} =$$

A. 1

B. 8/7

C. 7/5

D. 12/5

E. 5

Attachment:
tpr.png

$$\sqrt{a} = 3$$ and $$\sqrt{b} = 4$$, thus $$a= 9$$ and $$b = 16$$.

$$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a + b}}=\frac{3+4}{\sqrt{9+16}}=\frac{7}{5}$$

Hope it's clear.
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Re: If a^(1/2) = 3 and b^(1/2) = 4, then (a^(1/2) + b^(1/2)/(a + b)^(1/2)  [#permalink]

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21 Aug 2019, 19:22
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Re: If a^(1/2) = 3 and b^(1/2) = 4, then (a^(1/2) + b^(1/2)/(a + b)^(1/2)   [#permalink] 21 Aug 2019, 19:22
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