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If y^-2 + 2y^-1

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If y^-2 + 2y^-1 [#permalink] New post 02 Mar 2018, 16:29
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Question Stats:

53% (01:45) correct 46% (01:51) wrong based on 13 sessions
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. \(-5\)
[Reveal] Spoiler: OA

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Re: If y^-2 + 2y^-1 [#permalink] New post 02 Mar 2018, 16:39
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Explanation

\(\frac{1}{y^2} + \frac{1}{2y^1} -\frac{1}{15} = 0\)

This is like to write \(y^2 + 2y - 15 =0\)

Now, this is a quadratic equation, in which the multiplication is 15 and the sum is 2.

Two numbers that give us this are +5 and -3

y + 5 =0

y - 3 =0

OR

y = - 5
y = 3

At this point, the answer could be A and E but we do know that the answer must be one and only. Do not forget that we flipped the entire equation for simplicity. We should turn back because our y was raised to -1.

Which means that our 3 becomes \(\frac{1}{3}\) and our -5 become \(\frac{- 1}{5}\).

As you can see, we do have one and only correct solution among the answer choices that is \(\frac{- 1}{5}\).

C is correct
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Re: If y^-2 + 2y^-1 [#permalink] New post 30 Apr 2019, 16:24
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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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Re: If y^-2 + 2y^-1   [#permalink] 30 Apr 2019, 16:24
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