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If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu

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If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 05 Jul 2018, 21:28
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If \(y^{-2} + 2y^{-1} -15 = 0\), which of the following could be the value of y?


A. \(3\)

B. \(\frac{1}{5}\)

C. \(\frac{-1}{5}\)

D. \(\frac{-1}{3}\)

E. \(-5\)

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If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 05 Jul 2018, 21:38
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Bunuel wrote:
If \(y^{-2} + 2y^{-1} -15 = 0\), which of the following could be the value of y?


A. \(3\)

B. \(\frac{1}{5}\)

C. \(\frac{-1}{5}\)

D. \(\frac{-1}{3}\)

E. \(-5\)


Let \(y^{-1}=x\)

So, \(y^{-2} + 2y^{-1} -15 = 0\)
Or, \(x^2+2x-15=0\)
Or, x=-5 or x=3

1. x=-5, or, \(y^{-1}=-5\) or, \(y=\frac{-1}{5}\)
2. x=3,or, \(y^{-1}=3\) or, \(y=\frac{1}{3}\)

Among given options, correct answer is (C)
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Re: If y^(-2) + 2y^(-1) -15 = 0, which of the following could be the value  [#permalink]

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New post 08 Jul 2018, 05:26
Bunuel wrote:
If \(y^{-2} + 2y^{-1} -15 = 0\), which of the following could be the value of y?


A. \(3\)

B. \(\frac{1}{5}\)

C \(\frac{-1}{5}\)

D. \(\frac{-1}{3}\)

E. \(-5\)


Let \(y^{-1}=x\)

So, \(y^{-2} + 2y^{-1} -15 = 0\)
Or, \(x^2+2x-15=0\)
Or, x=-5 or x=3

1. x=-5, or, \(y^{-1}=-5\) or, \(y=\frac{-1}{5}\)
2. x=3,or, \(y^{-1}=3\) or, \(y=\frac{1}{3}\)

Among given options, correct answer is (C)[/quote]
P.S:- Already discussed .
https://gmatclub.com/forum/if-y-2-2y-1- ... 69804.html
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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 09 Jul 2018, 03:44
PKN wrote:
Bunuel wrote:
If \(y^{-2} + 2y^{-1} -15 = 0\), which of the following could be the value of y?


A. \(3\)

B. \(\frac{1}{5}\)

C. \(\frac{-1}{5}\)

D. \(\frac{-1}{3}\)

E. \(-5\)


Let \(y^{-1}=x\)

So, \(y^{-2} + 2y^{-1} -15 = 0\)
Or, \(x^2+2x-15=0\)
Or, x=-5 or x=3

1. x=-5, or, \(y^{-1}=-5\) or, \(y=\frac{-1}{5}\)
2. x=3,or, \(y^{-1}=3\) or, \(y=\frac{1}{3}\)

Among given options, correct answer is (C)

Is there an alternative way to solve this problem?
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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 09 Jul 2018, 03:51
Plugin answer choices and put in LHS of the given equation.The value of y for which the LHS becomes zero is the correct answer choice.

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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 22 Aug 2018, 06:54
1
\(y^{-2}\) can be written as \(\frac{1}{y^2}\)
Similarly \(y^{-1}\) can be written as \(\frac{1}{y}\)

Then the equation becomes \(\frac{1}{y^2}\) + 2\(\frac{1}{y}\) - 15 = 0.

Taking LCM, the equation becomes 1+2y-15\(y^2\) = 0.

Re-arranging gives 15\(y^2\)-2y-1 = 0.

15\(y^2\)-5y+3y-1 = 0.

5y(3y-1)+1(3y-1) = 0

Either y = \(\frac{-1}{3}\) or y = \(\frac{-1}{5}\).

Among the answer choices.

y = \(\frac{-1}{5}\)

C is the answer.
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If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 22 Aug 2018, 08:27
Afc0892

Sure it is a typo (since you got the answer) but -1/3 is not a root, otherwise option D will be correct too. The roots are 1/3 and -1/5 (option C). I am also an arsenal fan. Up gunners!!!


Afc0892 wrote:
\(y^{-2}\) can be written as \(\frac{1}{y^2}\)
Similarly \(y^{-1}\) can be written as \(\frac{1}{y}\)

Then the equation becomes \(\frac{1}{y^2}\) + 2\(\frac{1}{y}\) - 15 = 0.

Taking LCM, the equation becomes 1+2y-15\(y^2\) = 0.

Re-arranging gives 15\(y^2\)-2y-1 = 0.

15\(y^2\)-5y+3y-1 = 0.

5y(3y-1)+1(3y-1) = 0

Either y = \(\frac{-1}{3}\) or y = \(\frac{-1}{5}\).

Among the answer choices.

y = \(\frac{-1}{5}\)

C is the answer.
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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 22 Aug 2018, 08:32
The best way to go here is to put the options.
And only select options that is any option more than 1 will not satisfy.
Only options less than 1 will satisfy since the power is negative.

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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 22 Aug 2018, 17:55
funsogu wrote:
Afc0892

Sure it is a typo (since you got the answer) but -1/3 is not a root, otherwise option D will be correct too. The roots are 1/3 and -1/5 (option C). I am also an arsenal fan. Up gunners!!!


Afc0892 wrote:
\(y^{-2}\) can be written as \(\frac{1}{y^2}\)
Similarly \(y^{-1}\) can be written as \(\frac{1}{y}\)

Then the equation becomes \(\frac{1}{y^2}\) + 2\(\frac{1}{y}\) - 15 = 0.

Taking LCM, the equation becomes 1+2y-15\(y^2\) = 0.

Re-arranging gives 15\(y^2\)-2y-1 = 0.

15\(y^2\)-5y+3y-1 = 0.

5y(3y-1)+1(3y-1) = 0

Either y = \(\frac{-1}{3}\) or y = \(\frac{-1}{5}\).

Among the answer choices.

y = \(\frac{-1}{5}\)

C is the answer.


Yes its typo. Thanks for mentioning. COYG ?

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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu  [#permalink]

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New post 26 Aug 2018, 18:21
Bunuel wrote:
If \(y^{-2} + 2y^{-1} -15 = 0\), which of the following could be the value of y?


A. \(3\)

B. \(\frac{1}{5}\)

C. \(\frac{-1}{5}\)

D. \(\frac{-1}{3}\)

E. \(-5\)


Simplifying we have:

1/y^2 + 2/y - 15 = 0

Multiplying by y^2 we have:

1 + 2y - 15y^2 = 0

15y^2 - 2y - 1 = 0

(3y - 1)(5y + 1) = 0

3y - 1 = 0 → 3y = 1 → y = 1/3

Or

5y + 1 = 0 → 5y = -1 → y = -⅕

Answer: C
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Re: If y^(-2) + 2y^(-1) - 15 = 0, which of the following could be the valu &nbs [#permalink] 26 Aug 2018, 18:21
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