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Carcass
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a > 0 [#permalink] New post   21 Dec 2017, 16:53
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\(a > 0\)

Quantity A
Quantity B
\((a + a^{-1})^2\)
\(a^2 + a^{-2}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.

kudo for the right solution and explanation
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A
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Re: a > 0 [#permalink] New post   22 Dec 2017, 00:26
1
QA (A + A^-1)^2
A^2 + 2(A)(1/A) + (1/A)^2
A^2 + 2 + 1/A^2
QB
A^2 + A^-2 = A^2 + 1/A^2

Lets substract both side 1/A^2

QA A^2 +2
QB A^2
since we know that is positive, we can substract each side by A^2
QA = +2
QB = 0
So A

If we want to plu values, if this case we need to test fraction and interger positive right? (E.g. 1/2 and 2)
If this question did not mention anything about a, we could not do what I did?
How could we deal if we only had the two quantities and nothing about a, it could either -4, -(1/4), 1/4 and 4?
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Re: a > 0 [#permalink] New post   22 Dec 2017, 22:32
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Carcass wrote:
\(a > 0\)

Quantity A
Quantity B
\((a + a^{-1})^2\)
\(a^2 + a^{-2}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.

kudo for the right solution and explanation



From Statement 1 , we have

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


From Statement 2, we have

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

From the two statement we find out that Statement 1 has number 2 added to \((a^2 + \frac{1}{a^2})\)

so for any values of a>0 , Statement 1 will always be greater than statement 2

Hence Option - A
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Re: a > 0 [#permalink] New post   02 Jan 2022, 07:31
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Re: a > 0 [#permalink]
02 Jan 2022, 07:31
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