(Algebra) Simple question but I am confused.

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by cans » Wed Jun 08, 2011 9:36 pm
(x^2 + 4x + 4) / (x^2 - 4) = 3
(x+2)^2 / [(x-2)*(x+2) ] = 3
x is not equal to -2 (otherwise we will get 0/0)
thus (x+2)/(x-2)=3
solving, x=4
IMO A
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by blaster » Wed Jun 08, 2011 9:36 pm
we can simplify x^2+4x+4 to (x+2)^2

so

(x+2)^2/(x^2-4)=(x+2)^2/(x-2)*(x+2)= 3

after simplifing we get answer 4

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by smackmartine » Wed Jun 08, 2011 9:37 pm
IMO A

(x+2)^2/(x+2)(x-2) =3
(x+2)/(x-2) =3
(x+2)= 3x-6
2x=8
x=4

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by phanideepak » Wed Jun 08, 2011 9:38 pm
(x+2)^2/(x+2)(x-2) = 3

(x+2)/(x-2) = 3

solve for x and answer is 4

Did i miss something?

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by Anurag@Gurome » Wed Jun 08, 2011 9:39 pm
Rezinka wrote:Q
If (x^2 + 4x + 4) / (x^2 - 4) = 3 ; then x =

A. 4
B. 2
C. 0
D. -2
E. -4
(x² + 4x + 4) = (x + 2)²
(x² - 4) = (x - 2)(x + 2)
So, (x² + 4x + 4)/(x² - 4) = (x + 2)²/(x - 2)(x + 2) = (x + 2)/(x - 2)
So, (x + 2)/(x - 2) = 3
x = 2 = 3x - 6
8 = 2x
x = 4

The correct answer is A.
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by Rezinka » Wed Jun 08, 2011 9:40 pm
Yeah.. I had done the same thing. Made a stupid mistake. Silly me!
Actually, I missed that x^2 should not be 4 and not x. So eliminated option A.

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by Blimey72 » Thu Jun 09, 2011 2:30 am
Also try plugging in the ACs. Often the quickest and simplest method.

Can rule out B & D straightaway (as would otherwise be dividing by 0 which is not allowed).
x=0, means the fraction results in -1 (not 3 as needed), so C is out.
Then just plug in 4 or -4 and decide between them. All the numbers are fairly low, so can probably even do the math in your head.

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by [email protected] » Thu Jun 09, 2011 2:39 am
OA iS A...
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by [email protected] » Thu Jun 09, 2011 2:40 am
OA iS A...
IT IS TIME TO BEAT THE GMAT

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Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.

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by MBA.Aspirant » Thu Jun 09, 2011 8:51 pm
Rezinka wrote:Q
If (x^2 + 4x + 4) / (x^2 - 4) = 3 ; then x =

A. 4
B. 2
C. 0
D. -2
E. -4
What if you solved this way?

(x^2 + 4x + 4) / (x^2 - 4) = 3

x^2 +4x+4 = 3x^2 -12

2x^2 -4x-16=0

x^2 -2x-8=0

(x-4) (x+2)=0

x= 4 or x = -2

how can you choose then? I guess because -2 can't be in the denominator so it's out.

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by smackmartine » Thu Jun 09, 2011 9:43 pm
MBA.Aspirant wrote:
Rezinka wrote:Q
If (x^2 + 4x + 4) / (x^2 - 4) = 3 ; then x =

A. 4
B. 2
C. 0
D. -2
E. -4

What if you solved this way?

(x^2 + 4x + 4) / (x^2 - 4) = 3

x^2 +4x+4 = 3x^2 -12

2x^2 -4x-16=0

x^2 -2x-8=0

(x-4) (x+2)=0

x= 4 or x = -2

how can you choose then? I guess because -2 can't be in the denominator so it's out.
You need not have to stop at this point. In fact you should substitute each value of x into the orignal equation to check its validity.

at x= -2 ,

((-2)^2 + 4x + 4) / ((-2)^2 - 4) = 3 , is undefined as denominator becomes zero. This means x=-2 is not a valid value. Only at x=4 LHS = RHS (left hand side = right hand side)

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by vinayreguri » Sun Jun 12, 2011 5:44 am
Try plugging in Answer choices, Then Ans A