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Which of the following are roots of an equation (x^-2)+(2x^-1)-15=0 A.
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02 Mar 2021, 09:11
02 Mar 2021, 09:11
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Which of the following are roots of an equation \((x^{-2})+(2x^{-1})-15=0\)
A. 1/5 and -1/3
B. -1/5 and 1/3
C. 1/5 and 1/3
D. -1/5 and -1/3
E. -5/2 and -1/3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A. 1/5 and -1/3
B. -1/5 and 1/3
C. 1/5 and 1/3
D. -1/5 and -1/3
E. -5/2 and -1/3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Re: Which of the following are roots of an equation (x^-2)+(2x^-1)-15=0 A.
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02 Mar 2021, 09:34
02 Mar 2021, 09:34
2
Take \(x^{(-1)}\) = y
So the equation will be y^2 + 2y - 15 = 0
By solving, (y+5)(y-3) = 0
y = -5 OR 3
\(x^{(-1)}\) = y
So x = \(\frac{1}{y}\)
x = \(\frac{-1}{5}\) OR \(\frac{1}{3}\)
Answer B
So the equation will be y^2 + 2y - 15 = 0
By solving, (y+5)(y-3) = 0
y = -5 OR 3
\(x^{(-1)}\) = y
So x = \(\frac{1}{y}\)
x = \(\frac{-1}{5}\) OR \(\frac{1}{3}\)
Answer B
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Re: Which of the following are roots of an equation (x^-2)+(2x^-1)-15=0 A.
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02 Mar 2021, 09:49
02 Mar 2021, 09:49
1
Carcass wrote:
Which of the following are roots of an equation \((x^-2)+(2x^-1)-15=0\)
A. 1/5 and -1/3
B. -1/5 and 1/3
C. 1/5 and 1/3
D. -1/5 and -1/3
E. -5/2 and -1/3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A. 1/5 and -1/3
B. -1/5 and 1/3
C. 1/5 and 1/3
D. -1/5 and -1/3
E. -5/2 and -1/3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Given: \(x^{-2} + 2x^{-1} - 15 = 0\)
Rewrite as: \(\frac{1}{x^2} + \frac{2}{x} - 15 = 0\)
Eliminate fractions by multiplying both sides by \(x^2\) to get: \(1 + 2x - 15x^2 = 0\)
Rearrange to get: \(15x^2 - 2x - 1 = 0\)
Factor to get: \((5x + 1)(3x - 1) = 0\)
So, EITHER \(5x + 1 = 0\) OR \(3x - 1 = 0\)
If \(5x + 1 = 0\), then \(x = -\frac{1}{5}\)
If \(3x - 1 = 0\), then \(x = \frac{1}{3}\)
So, the roots (solutions) are \(-\frac{1}{5}\) and \(\frac{1}{3}\)
Answer: B
