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# What is the value of 5x^2 + 4x-1

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Senior Manager
Joined: 02 Sep 2006
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What is the value of 5x^2 + 4x-1 [#permalink]

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01 Mar 2007, 01:39
00:00

Difficulty:

15% (low)

Question Stats:

76% (00:37) correct 24% (00:45) wrong based on 226 sessions

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What is the value of 5x^2 + 4x-1

(1) x(x+2)=0
(2) x=0
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Mar 2012, 12:03, edited 2 times in total.
Edited the question

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SVP
Joined: 01 May 2006
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01 Mar 2007, 01:51
(B) for me

5*x^2+ 4x-1 = ?

From 1
x*(x+2) =0
<=> x=0 or x=-2

Let's plug the values to see if the result would be equal:
o If x=0, then 5*0 + 4*0 -1 = -1
o If x=-2, then 5*4 + 4*(-2) -1 = 20 - 9 = 11

INSUFF.

From 2
x=0

Clearly, we have : 5*0 + 4*0 -1 = -1

SUFF

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Intern
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WE: Other (Transportation)
Re: What is the value of 5(x)2+ 4x-1 1) x(x+2)=0 2)x=0 [#permalink]

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10 Mar 2012, 10:57
1. x(x+2)=0 <=> x=0 or x=-2 Insuff (We need only one value for x)
2. x=0 Clearly sufficient. Replace x with 0 and you have the value.

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Senior Manager
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Joined: 05 Sep 2016
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Re: What is the value of 5x^2 + 4x-1 [#permalink]

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17 Sep 2016, 07:42
B is correct

(1) x(x+2) = 0 --> x = 0 or x = -2

NOT SUFFICIENT - we do not have a definitive answer for the value of x to plug into the main eq

(2) x = 0

SUFFICIENT - plug straight into main eq

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Director
Joined: 26 Oct 2016
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Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
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WE: Education (Education)
Re: What is the value of 5x^2 + 4x-1 [#permalink]

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26 Jan 2017, 19:59
It's pretty straight forward

St1 :-
directly gives two values of x which itself makes the statement insufficient.

St2 :-
We are directly given value of x which can be used to calculate the value of the function.
Suffi.

Hence B.
_________________

Thanks & Regards,
Anaira Mitch

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Intern
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Re: What is the value of 5x^2 + 4x-1 [#permalink]

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29 Jan 2017, 10:30
Statement 1: It is a quadratic equation which gives two values of x ; 0 and -2 . Hence, statement is Insufficient
Statement 2 : Direct value of X=0 gives the result. SUFFICIENT

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Re: What is the value of 5x^2 + 4x-1   [#permalink] 29 Jan 2017, 10:30
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