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x2 + 1x2 = ?

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x2 + 1x2 = ? [#permalink]

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\(x^2 + \frac{1}{x^2}\) = ?

1) \(x - \frac{1}{x} = 4\)
2) \(x + \frac{1}{x} = 2√5\)
[Reveal] Spoiler: OA

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x2 + 1x2 = ? [#permalink]

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New post 29 Mar 2017, 01:12
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MathRevolution wrote:
\(x^2 + \frac{1}{x^2}\) = ?

1) \(x - \frac{1}{x} = 4\)
2) \(x + \frac{1}{x} = 2√5\)


St I
\(x - \frac{1}{x} = 4\)

on squaring both sides, we have

\((x - \frac{1}{x})^2 = 4^2\)

\(x^2 + \frac{1}{x^2} - 2 = 16\)

\(x^2 + \frac{1}{x^2} = 18\) ----------------Sufficient


St II
\(x + \frac{1}{x} = 2√5\)

on squaring both sides, we have

\((x + \frac{1}{x})^2 = (2√5)^2\)

\(x^2 + \frac{1}{x^2} + 2 = 20\)

\(x^2 + \frac{1}{x^2} = 18\) ----------------Sufficient

Hence option D is correct
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Re: x2 + 1x2 = ? [#permalink]

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New post 31 Mar 2017, 00:33
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==> In the original condition, there is 1 variable (x) and in order to match the number of variables to the number of equations, there must be 1 equation. Since there is 1 for con 1) and 1 for con 2), D is most likely to be the answer. If you actually solve this, you get con 1) = con 2), and thus if you square both sides,
from \(x^2 + \frac{1}{x^2} -2=16\) and \(x^2 + \frac{1}{x^2} +2=20\), you get \(x^2 + \frac{1}{x^2} =18\), hence unique and sufficient.

Therefore, the answer is D.
Answer: D
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Re: x2 + 1x2 = ? [#permalink]

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New post 22 May 2017, 14:53
MathRevolution wrote:
\(x^2 + \frac{1}{x^2}\) = ?

1) \(x - \frac{1}{x} = 4\)
2) \(x + \frac{1}{x} = 2√5\)


This question appears to be very simple if you know the properties of foiling; however, this is a question designed to bait the test taker into picking the classic "C" both are sufficient trap. While it knowing (x-y) and (x+y) would be sufficient to solve the question, it is actually not necessary and a skill the GMAT is testing you on... that the GMAT is just testing high school math and algebra is a myth because this clearly this question shows how arithmetic and manual become derelict when calculators are used unnecessarily - well in Singapore it really is but that's another story. Anyways,

Statement 1

\(x - \frac{1}{x} = 4\) square both sides
\(x - \frac{1}{x} [m]x - \frac{1}{x} = 16
x^2- x(1/x)-x(1/x) + (1/x^2) = 16
x^2-2x(1/x) + (1/x^2) =16 ( pay attention to the reciprocal property)
x^2- 2 +(1/x^2) = 16
x^2 + (1/x^2) = 18

Statement 2

[m]x + \frac{1}{x} = 2√5\)
[m]x + \frac{1}{x} [m]x + \frac{1}{x} = 20
x^2 + x(1/x) + x(1/x) + 1/x^2 = 20
x^2 + 2x (1/x) + (1/x^2) = 20 (pay close attention to the reciprocal property- x of any number times its reciprocal is 1 so 2x times the reciprocal of just x is always 2)
x^2 + 2 + (1/x^2) = 20
x^2 + (1/x^2) = 18

Lastly- it is important to note that even though statement 1 and 2 both reduced to a sum of 18- they don't necessarily have to have the same sum. For example, if statement 1 allows you to find x^2 + (1/x^2) and statement 2 also allows you to find x^2 + (1/x^2) but happens to have a different result, say 16 instead of 18, it would still be D.

Thus "D" is the correct answer
Re: x2 + 1x2 = ?   [#permalink] 22 May 2017, 14:53
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