Carcass 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's use some
u-substitution!
Notice that we can write \(y^{-2}\) and \(2y^{-1}\) as powers of \(\frac{1}{y}\)
Here's what I mean:
\(y^{-2} = \frac{1}{y^2}=(\frac{1}{y})^2\)
and
\(y^{-1} = \frac{1}{y^1}=(\frac{1}{y})^1=\frac{1}{y}\)
So, if we let \(u = \frac{1}{y}\), we can rewrite the equation as follows: \(u^2 + 2u -15 = 0\)
Factor: \((u+5)(u-3)=0\)
So, EITHER \(u = -5\) OR \(u = 3\)
If \(u = -5\), then \(\frac{1}{u} = -5\), which means \(u = \frac{-1}{5}\)
If \(u = 3\), then \(\frac{1}{u} = 3\), which means \(u = \frac{1}{3}\)
Check the answer choices . . .
Answer: C
Cheers,
Brent
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Brent Hanneson - founder of Greenlight Test PrepSign up for our GRE Question of the Day emails