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

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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