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M28-15

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M28-15  [#permalink]

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New post 16 Sep 2014, 00:28
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

75% (00:34) correct 25% (00:49) wrong based on 24 sessions

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Re M28-15  [#permalink]

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New post 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 \(12-2\sqrt{35}\approx{12-2*6}=0\).


Answer: A
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Re: M28-15  [#permalink]

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New post 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.
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Re: M28-15  [#permalink]

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New post 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



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Re: M28-15 &nbs [#permalink] 18 Dec 2017, 08:38
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