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

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Bunuel
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(a^4 + 1)/a^2 = [#permalink] New post   15 Nov 2019, 01:42
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A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

68% (02:15) correct 32% (02:08) wrong based on 224 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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E
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Re: (a^4 + 1)/a^2 = [#permalink] New post   15 Nov 2019, 01:52
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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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Re: (a^4 + 1)/a^2 = [#permalink] New post   22 Nov 2019, 08:06
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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:)
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philipssonicare
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Re: (a^4 + 1)/a^2 = [#permalink] New post   11 Apr 2021, 18:21
I eliminated A, C, and D as they are similar to difference of squares.

Tested B and it didn't work. Weak, but gets to the answer E

Dillesh4096 wrote:
\(\frac{a^4 + 1}{a^2}\)
--> \(a^2 + \frac{1}{a^2}\)
--> \(a^2 + \frac{1}{a^2} + 2*a*\frac{1}{a} - 2*a*\frac{1}{a}\)


I don't understand how I'm meant to see to do this looking at the equation.
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Re: (a^4 + 1)/a^2 = [#permalink] New post   06 Oct 2023, 00:55
philipssonicare wrote:
I eliminated A, C, and D as they are similar to difference of squares.

Tested B and it didn't work. Weak, but gets to the answer E

Dillesh4096 wrote:
\(\frac{a^4 + 1}{a^2}\)
--> \(a^2 + \frac{1}{a^2}\)
--> \(a^2 + \frac{1}{a^2} + 2*a*\frac{1}{a} - 2*a*\frac{1}{a}\)


I don't understand how I'm meant to see to do this looking at the equation.



You can only if you able to calculate the next steps vaguely in your head before you write it down.
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Re: (a^4 + 1)/a^2 = [#permalink]
06 Oct 2023, 00:55
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