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# (a^4 + 1)/a^2 =

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Math Expert
Joined: 02 Sep 2009
Posts: 59622

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15 Nov 2019, 01:42
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Difficulty:

55% (hard)

Question Stats:

65% (02:10) correct 35% (01:39) wrong based on 40 sessions

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$$\frac{a^4 + 1}{a^2}$$ =

A. $$(a+\frac{1}{a})(a-\frac{1}{a})$$

B. $$(a+\frac{1}{a})^2$$

C. $$(a + \frac{\sqrt{2}}{a})(a - \frac{\sqrt{2}}{a})$$

D. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} - \frac{1}{a})$$

E. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} + \frac{1}{a})$$

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Joined: 20 Jul 2017
Posts: 1143
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: (a^4 + 1)/a^2 =  [#permalink]

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15 Nov 2019, 01:52
Bunuel wrote:
$$\frac{a^4 + 1}{a^2}$$ =

A. $$(a+\frac{1}{a})(a-\frac{1}{a})$$

B. $$(a+\frac{1}{a})^2$$

C. $$(a + \frac{\sqrt{2}}{a})(a - \frac{\sqrt{2}}{a})$$

D. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} - \frac{1}{a})$$

E. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} + \frac{1}{a})$$

$$\frac{a^4 + 1}{a^2}$$
--> $$a^2 + \frac{1}{a^2}$$
--> $$a^2 + \frac{1}{a^2} + 2*a*\frac{1}{a} - 2*a*\frac{1}{a}$$
--> $$(a + \frac{1}{a})^2 - 2$$
--> $$(a + \frac{1}{a})^2 - (\sqrt{2})^2$$
--> $$(a + \frac{1}{a} + \sqrt{2})(a + \frac{1}{a} - \sqrt{2})$$

IMO Option E
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Joined: 10 Dec 2017
Posts: 151
Location: India
Re: (a^4 + 1)/a^2 =  [#permalink]

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22 Nov 2019, 08:06
Bunuel wrote:
$$\frac{a^4 + 1}{a^2}$$ =

A. $$(a+\frac{1}{a})(a-\frac{1}{a})$$

B. $$(a+\frac{1}{a})^2$$

C. $$(a + \frac{\sqrt{2}}{a})(a - \frac{\sqrt{2}}{a})$$

D. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} - \frac{1}{a})$$

E. $$(a + \sqrt{2} + \frac{1}{a})(a - \sqrt{2} + \frac{1}{a})$$

Are You Up For the Challenge: 700 Level Questions

Plug a=1
Required value=2
Check options
E gives 2
E:)
Re: (a^4 + 1)/a^2 =   [#permalink] 22 Nov 2019, 08:06
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