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The values of x for which the expression above is NOT defined are

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Bunuel
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The values of x for which the expression above is NOT defined are [#permalink] New post  22 May 2020, 07:53
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\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2


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GMATinsight
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  15 Jun 2020, 05:40
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Kritisood wrote:
GMATinsight

\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

if i solve this further i get

\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)

cancelling like terms

\(\frac{1}{x-1} - \frac{1}{x-2}\)

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?


Kritisood

Canceling Like terms is acceptable ONLY

- If the term to be canceled is NON-ZERO

You are canceling the term (x+1) without knowing whether it's NON-ZERO - \(This is the mistake\)

Since that term also may be 0 bring the situation of \(\frac{0}{0}\) therefore it will be WRONG to cancel it.

SImilar example
\(x^2-2x = 0\) does NOT mean \(x*x = 2x\) and \(x = 2 \)
\(x^2-2x = 0\) means \(x*(x-2) = 0\) i.e. x = 0 or 2

I hope this help! :)
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  22 May 2020, 08:11
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Bunuel wrote:
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2


PS21172


Key concept: A rational expression (aka fraction) is undefined when the denominator equal zero

So let's check each denominator individually and see what values of x make that denominator equal to zero.

If \(x^2 - 1 = 0\), then \(x^2 = 1\), which means \(x=1\) and \(x=-1\)
In other words, the expression is not defined when \(x=1\) and when \(x=-1\)

If \(x^2 - 4x + 4 = 0\), then we can factor the left side to get: \((x-2)(x-2)= 0\)
This means \(x = 2\)
So, the expression is not defined when \(x=2\)

So the expression is undefined when \(x=1\), \(x=-1\) and \(x=2\)

Answer: E

Cheers,
Brent
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  22 May 2020, 09:25
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Bunuel wrote:
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2


PS21172


CONCEPT: an expression is NOT DEFINED when denominator = 0


i.e. in the expression \(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

\(x^2 - 1 = 0\) i.e. x = +1 or -1
and
\(x^2 - 4x + 4 = 0\) i.e \((x -2)^2 = 0\) i.e. x = 2

Answer: Option E
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The values of x for which the expression above is NOT defined are [#permalink] New post  15 Jun 2020, 05:09
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GMATinsight

\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

if i solve this further i get

\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)

cancelling like terms

\(\frac{1}{x-1} - \frac{1}{x-2}\)

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?
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Kritisood
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  15 Jun 2020, 07:35
GMATinsight wrote:
Kritisood wrote:
GMATinsight

\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

if i solve this further i get

\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)

cancelling like terms

\(\frac{1}{x-1} - \frac{1}{x-2}\)

now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?


Kritisood

Canceling Like terms is acceptable ONLY

- If the term to be canceled is NON-ZERO

You are canceling the term (x+1) without knowing whether it's NON-ZERO - \(This is the mistake\)

Since that term also may be 0 bring the situation of \(\frac{0}{0}\) therefore it will be WRONG to cancel it.

SImilar example
\(x^2-2x = 0\) does NOT mean \(x*x = 2x\) and \(x = 2 \)
\(x^2-2x = 0\) means \(x*(x-2) = 0\) i.e. x = 0 or 2

I hope this help! :)


This helps a lot, thank you so much GMATinsight
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Basshead
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  27 Nov 2020, 14:05
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Bunuel wrote:
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

The values of x for which the expression above is NOT defined are

A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2


PS21172


\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

\(\frac{x+1}{(x+1)(x-1)} - \frac{x-2 }{ (x-2)(x-2)}\)

An expression is undefined if the denominator will equal 0.

From the above, the expression will equal 0 is \(x = -1, 1, or 2\). Answer is E.

Notice we CAN'T cancel out \(x+1\) on the left hand side because we could be canceling out a root.

Answer is E.
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Re: The values of x for which the expression above is NOT defined are [#permalink] New post  06 Mar 2023, 09:41
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)

The values of x for which the expression above is NOT defined are

Since denominators contain terms e.g. (x-1), (x+1) & (x-2)
The values of x for which the expression above is NOT defined are 1, -1 & 2

IMO E
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Re: The values of x for which the expression above is NOT defined are [#permalink]
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