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If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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19 Jan 2012, 12:47
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If x is not equal to zero, and \(x + \frac{1}{x} = 3\), then what is the value of \(x^4 + (\frac{1}{x})^4\)? A. 27 B. 32 C. 47 D. 64 E. 81
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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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23 Jan 2012, 22:33
we can try it with the following method as well x + 1/x=3 we square both sides so we have x^2 + 1/x^2 +2 = 9 or x^2 + 1/x^2= 7
squaring again we have x^4 + 1/x^4 + 2 = 49 or x^4 + 1/x^4 = 47 answer =47 (C)




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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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19 Jan 2012, 13:51
Splendidgirl666 wrote: If x is not equal to zero, and x+1/x = 3, then what is the value of x^4 + (1/x)^4?
1. 27 2. 32 3. 47 4. 64 5. 81 I'd solve this one with approximation method rather than solving for x (which will give some irrational number) or squaring twice. \(x+\frac{1}{x}=3\) > \(x\approx{\frac{5}{2}}\) > \(\frac{5}{2}+\frac{2}{5}=2.5+0.4=2.9\) (so x is a little bit more than 5/2). \(x^4+\frac{1}{x^4}=\frac{625}{16}+\frac{16}{625}\) > \(\frac{16}{625}\) is a negligible value and \(\frac{625}{16}\approx{40}\). So as x is a little bit more than 5/2 then the answer should be a little bit more than 40 (taking into account the magnitude of 4th power). Best option is 47. Answer: C.
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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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19 Jan 2012, 15:06
Thanks. But if you approx as x=2.5 and 1/x as 0.5 then x^4+ 1/x^4 gives around 36. So why would I chose 47 as answer rather than 32?
What would be your tip for approx method here?



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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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19 Jan 2012, 15:21



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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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24 Jan 2012, 00:21
I find puneets method much simpler. I don't think it's intuitive but it's much simpler. Secondly, what is the source of this question? Posted from GMAT ToolKit



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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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24 Jan 2012, 01:48
puneetkr wrote: we can try it with the following method as well x + 1/x=3 we square both sides so we have x^2 + 1/x^2 +2 = 9 or x^2 + 1/x^2= 7
squaring again we have x^4 + 1/x^4 + 2 = 49 or x^4 + 1/x^4 = 47 answer =47 (C) scbguy wrote: I find puneets method much simpler. I don't think it's intuitive but it's much simpler. Secondly, what is the source of this question? As I've mentioned squaring twice is one of the ways to solve and it's actually quite easy and straightforward as puneetkr showed: Square once: \(x^2+2*x*\frac{1}{x}+\frac{1}{x^2}=9\) > \(x^2+\frac{1}{x^2}=7\); Square again: \(x^4+2*x^2*\frac{1}{x^2}+\frac{1}{x^4}=49\) > \(x^4+\frac{1}{x^4}=47\). Answer: C. The point is that number 5/2 poppedup in my mind in 2 seconds and taking it to the fourth power took another 2 seconds (5^4=625 and 2^4=16), so I arrived to the correct answer in less than 10 seconds. Though if this doesn't work for you, then algebraic way is also perfectly OK and not lengthy at all.
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Re: If x is not equal to zero, and x + 1/x = 3, then what is the value of [#permalink]
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18 Oct 2016, 09:39
Splendidgirl666 wrote: If x is not equal to zero, and x+1/x = 3, then what is the value of x^4 + (1/x)^4?
1. 27 2. 32 3. 47 4. 64 5. 81 \(x+\frac{1}{x} = 3\) \((x+\frac{1}{x})^2\) \(= x^2 + \frac{1}{x^2} + 2 = 9\) So, \(x^2 + \frac{1}{x^2} = 7\) Now, \((x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2 = 7^2\) Or, \(x^4 + \frac{1}{x^4} + 2 = 49\) Or, \(x^4 + \frac{1}{x^4} = 47\) Hence correct answer will be (C)
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