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# S95-26

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Math Expert
Joined: 02 Sep 2009
Posts: 55263
S95-26  [#permalink]

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16 Sep 2014, 01:50
00:00

Difficulty:

45% (medium)

Question Stats:

59% (01:17) correct 41% (01:43) wrong based on 39 sessions

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$$(\sqrt{7 + \sqrt{48}} + \sqrt{7 - \sqrt{48}})^2 =$$

A. $$1$$
B. $$7 - 4\sqrt{3}$$
C. $$14 - 4\sqrt{3}$$
D. $$14$$
E. $$16$$

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Math Expert
Joined: 02 Sep 2009
Posts: 55263
Re S95-26  [#permalink]

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16 Sep 2014, 01:50
Official Solution:

$$(\sqrt{7 + \sqrt{48}} + \sqrt{7 - \sqrt{48}})^2 =$$

A. $$1$$
B. $$7 - 4\sqrt{3}$$
C. $$14 - 4\sqrt{3}$$
D. $$14$$
E. $$16$$

The question asks for the value of an expression. This expression is in the form of $$(x + y)^2$$, where

$$x = \sqrt{7 + \sqrt{48}}$$ and $$y = \sqrt{7 - \sqrt{48}}$$.

The expanded form of $$(x + y)^2$$ is $$x^2 + 2xy + y^2$$. If we substitute our values for $$x$$ and $$y$$, we get:

$$(\sqrt{7 + \sqrt{48}})^2 + 2(\sqrt{7 + \sqrt{48}})(\sqrt{7 - \sqrt{48}}) + (\sqrt{7 - \sqrt{48}})^2$$

We simplify, recalling that the product of two radical terms can be rewritten under one radical sign:

$$(7 + \sqrt{48}) + 2\leftarrow(\sqrt{(7 + \sqrt{48})(7 - \sqrt{48})}\rightarrow) + (7 - \sqrt{48})$$.

The middle term is the factored form of the difference of two squares, $$(x + y)(x - y) = x^2 - y^2$$. Simplify accordingly:

$$(7 + \sqrt{48}) + 2\leftarrow(\sqrt{(49 - 48)}\rightarrow) + (7 - \sqrt{48})$$

Simplify further, noting that the $$\sqrt{48}$$ terms cancel:

$$7 + 7 + 2\sqrt{1} = 14 + 2 = 16$$

Answer: E
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Joined: 10 Jul 2017
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Re: S95-26  [#permalink]

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31 Aug 2017, 07:18
Very good question
Re: S95-26   [#permalink] 31 Aug 2017, 07:18
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# S95-26

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