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If y^(2) + 2y^(1)  15 = 0, which of the following could be the valu
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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
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05 Jul 2018, 21:38
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, \(x^2+2x15=0\) 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
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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, \(x^2+2x15=0\) 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/ify22y1 ... 69804.html
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Re: If y^(2) + 2y^(1)  15 = 0, which of the following could be the valu
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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, \(x^2+2x15=0\) 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
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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. Posted from my mobile device
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Re: If y^(2) + 2y^(1)  15 = 0, which of the following could be the valu
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22 Aug 2018, 06:54
\(y^{2}\) 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+2y15\(y^2\) = 0. Rearranging gives 15\(y^2\)2y1 = 0. 15\(y^2\)5y+3y1 = 0. 5y(3y1)+1(3y1) = 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
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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+2y15\(y^2\) = 0.
Rearranging gives 15\(y^2\)2y1 = 0.
15\(y^2\)5y+3y1 = 0.
5y(3y1)+1(3y1) = 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
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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
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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+2y15\(y^2\) = 0.
Rearranging gives 15\(y^2\)2y1 = 0.
15\(y^2\)5y+3y1 = 0.
5y(3y1)+1(3y1) = 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 ? Posted from my mobile device
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Re: If y^(2) + 2y^(1)  15 = 0, which of the following could be the valu
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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
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