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# M28-15

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Math Expert
Joined: 02 Sep 2009
Posts: 53020

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16 Sep 2014, 00:28
00:00

Difficulty:

35% (medium)

Question Stats:

76% (00:34) correct 24% (00:49) wrong based on 25 sessions

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If $$x=(\sqrt{5}-\sqrt{7})^2$$, then the best approximation of $$x$$ is:

A. $$0$$
B. $$1$$
C. $$2$$
D. $$3$$
E. $$4$$

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Math Expert
Joined: 02 Sep 2009
Posts: 53020

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16 Sep 2014, 00:29
Official Solution:

If $$x=(\sqrt{5}-\sqrt{7})^2$$, then the best approximation of $$x$$ is:

A. $$0$$
B. $$1$$
C. $$2$$
D. $$3$$
E. $$4$$

$$x=(\sqrt{5}-\sqrt{7})^2=5-2\sqrt{35}+7=12-2\sqrt{35}$$.

Since $$\sqrt{35}\approx{6}$$, then $$12-2\sqrt{35}\approx{12-2*6}=0$$.

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Joined: 07 Jun 2017
Posts: 1

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18 Dec 2017, 07:54
Please could you elaborate on the solution?

I get that squaring the current equation will produce 5-7 but I do not understand where the 2xroot32 comes from. Is that not another way of producing root7 squared - which is already accounted for in the equation.
Math Expert
Joined: 02 Sep 2009
Posts: 53020

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18 Dec 2017, 08:38
KarimAmer wrote:
Please could you elaborate on the solution?

I get that squaring the current equation will produce 5-7 but I do not understand where the 2xroot32 comes from. Is that not another way of producing root7 squared - which is already accounted for in the equation.

(a - b)^2 = a^2 - 2ab + b^2.

Before attempting questions one should brush-up fundamentals.

7. Algebra

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: M28-15   [#permalink] 18 Dec 2017, 08:38
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# M28-15

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