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

Question

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

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

GMATinsight · Accepted Answer

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 = 2 x and x = 2 x^2-2x = 0 means x*(x-2) = 0 i.e. x = 0 or 2 I hope this help!

BrentGMATPrepNow · Answer

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

GMATinsight · Answer

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

Kritisood · Answer

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?

Basshead · Answer

\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.

Kinshook · Answer

\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

DanTheGMATMan · Answer

Be careful about cancelling (x+1):

https://www.youtube.com/watch?v=usIGC-YefgQ

ablatt4 · Answer

If a function is undefined, it means that ANY denominator is 0.
 
There are 2 denominators here, we need to set each = 0

x^2-1=0
x^2=1
x=1,-1

x^2-4x-4=0

(x-2)^2=0

x=2

2,-1,1 make the function undefined

E

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

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

Prep Toolkit

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