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# 1/(4 - 15^(1/2))^2

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Math Expert
Joined: 02 Sep 2009
Posts: 57191

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23 Jul 2018, 23:27
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Difficulty:

25% (medium)

Question Stats:

69% (01:35) correct 31% (01:18) wrong based on 104 sessions

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$$(\frac{1}{4- \sqrt{15}})^2$$

A. $$16 + 16\sqrt {15}$$

B. $$31 - 8\sqrt {15}$$

C. $$31 + 8\sqrt {15}$$

D. $$32 - 4\sqrt {15}$$

E. $$32 + 4\sqrt {15}$$

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3018

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23 Jul 2018, 23:36

Solution

To find:
• The value of the expression $$(\frac{1}{4- \sqrt{15}})^2$$

Approach and Working:
We can rewrite the expression as:
• $$(\frac{1}{4- \sqrt{15}})^2$$
= $$({4+ \sqrt{{15}}})^2$$
= $$16 + 15 + 2 * 4 * \sqrt{{15}}$$
= $$31 + 8\sqrt{{15}}$$

Hence, the correct answer is option C.

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Math Expert
Joined: 02 Aug 2009
Posts: 7755

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23 Jul 2018, 23:40
Bunuel wrote:
$$(\frac{1}{4- \sqrt{15}})^2$$

A. $$16 + 16\sqrt {15}$$

B. $$31 - 8\sqrt {15}$$

C. $$31 + 8\sqrt {15}$$

D. $$32 - 4\sqrt {15}$$

E. $$32 + 4\sqrt {15}$$

Whenever you see a something like $$4-\sqrt{15}$$ in denominator, remember the formula $$a^2-b^2=(a-b)(a+b)$$

$$(\frac{1}{4- \sqrt{15}})^2=(\frac{4+\sqrt{15}}{(4+\sqrt{15})(4- \sqrt{15})})^2=(\frac{4+\sqrt{15}}{4^2-(\sqrt{15})^2})^2=(\frac{4+\sqrt{15}}{16-15})^2=(4+\sqrt{15})^2=16+8\sqrt{15}+15=31+\sqrt{15}$$

C
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Joined: 18 Jun 2018
Posts: 267

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23 Jul 2018, 23:44
Bunuel wrote:
$$(\frac{1}{4- \sqrt{15}})^2$$

A. $$16 + 16\sqrt {15}$$

B. $$31 - 8\sqrt {15}$$

C. $$31 + 8\sqrt {15}$$

D. $$32 - 4\sqrt {15}$$

E. $$32 + 4\sqrt {15}$$

OA:C

$$(\frac{1}{4- \sqrt{15}})^2$$
$$=({{\frac{1}{4- \sqrt{15}}*{\frac{4+\sqrt{15}}{4+ \sqrt{15}}})^2$$

Using $$a^2 - b^2 =(a-b)(a+b)$$
$$=({\frac{4+\sqrt{15}}{16 - 15})^2$$
$$=({4+\sqrt{15})^2$$
$$=16+15 +2*4*\sqrt{15}$$ (Using $$(a+b)^2 =a^2+b^2 + 2ab$$)
$$=31 +8\sqrt{15}$$
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04 Aug 2019, 10:08
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Re: 1/(4 - 15^(1/2))^2   [#permalink] 04 Aug 2019, 10:08
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