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aliasjit
Is x = 3?

(1) x^2 + 6x + 9 = 0
(2) x^2 + 7x - 21 > 0

Stmt 1: (x+3)^2=0
x = +/- 3 NS

Stmnt 2 (x+7)(X-3)>0
x >-7 or x>3 NS

1+2 : x=-3 Suff.

Option C

Note that the answer to this question is A, not C.

(1) \(x^2 + 6x + 9 = 0\) --> \((x+3)^2 =0\) --> \(x = -3\), so we have a NO answer to the question. Sufficient.
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aliasjit
Stmt 1: (x+3)^2=0
x = +/- 3 NS

Stmnt 2 (x+7)(X-3)>0
x >-7 or x>3 NS

1+2 : x=-3 Suff.

Option C

Is x = 3?

(1) x^2 + 6x + 9 = 0
(2) x^2 + 7x - 21 > 0

Be careful-

If we factor statement we can see that -3 is our answer - (x+3)(x+3) - you take the opposite of the number inside the parentheses. You could also plug in 3 into the equation in statement 1 which would equal 36 which is not 0. A common GMAT trap, perhaps deliberately designed by the test makers, is getting the test taker to the assume of the truth of a statement in the stimulus (part of question that is not statement 1 or 2) that he or she is actually supposed to falsify/validate with regards to the information. Anyways, if we plug in 3 into statement then the resulting number will be greater than 0... but is this the only number that can satisfy the equation- if x is 4 then that also results in a number greater than 0.
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Hi ,

I know it is a silly question , But please explain how to factor the below equation or statement 2

x2+7x−21>0
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abhisheknandy08
Hi ,

I know it is a silly question , But please explain how to factor the below equation or statement 2

x2+7x−21>0

You cannot factor this equation- to get the roots of this equation you would have to use the quadratic formula- but that's not the point because the GMAT isn't really testing your ability to apply the quadratic- you do not need the roots of this equation to solve the problem
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abhisheknandy08
Hi ,

I know it is a silly question , But please explain how to factor the below equation or statement 2

x2+7x−21>0

You cannot factor this equation- to get the roots of this equation you would have to use the quadratic formula- but that's not the point because the GMAT isn't really testing your ability to apply the quadratic- you do not need the roots of this equation to solve the problem
And it's actually a good question- because even though the answer to it is not necessary to solve the problem sometimes you only arrive at that realization after you have your question answered and as such can rationalize both the question and an efficient method for yourself.
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Nunuboy1994
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abhisheknandy08
Hi ,

I know it is a silly question , But please explain how to factor the below equation or statement 2

x2+7x−21>0

You cannot factor this equation- to get the roots of this equation you would have to use the quadratic formula- but that's not the point because the GMAT isn't really testing your ability to apply the quadratic- you do not need the roots of this equation to solve the problem. And it's actually a good question- because even though the answer to it is not necessary to solve the problem sometimes you only arrive at that realization after you have your question answered and as such can rationalize both the question and an efficient method for yourself. If you look at Bunuel's explanations for even 700 questions, you'll see he never really uses anymore than 5 steps.
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abhisheknandy08
Hi ,

I know it is a silly question , But please explain how to factor the below equation or statement 2

x2+7x−21>0

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm
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Bunuel
Is x = 3?

(1) \(x^2 + 6x + 9 = 0\)

(2) \(x^2 + 7x - 21 > 0\)


Given : nothing
DS: x=3

Statement 1 : (1) \(x^2 + 6x + 9 = 0\)
\((x+3)^2 = 0\)
x= -3
SUFFICIENT

Statement 2 :

\(x^2 + 7x - 21 > 0\)

\(x^2 + 2*(7/2)x + 49/4 - 49/4 - 21 > 0\)
\((x+7/2)^2 - 133/4 > 0\)
\((x+7/2)< - \sqrt{133/4} or (x+7/2)> \sqrt{133/4}\)
\(x<-7/2- \sqrt{133/4} or (x>-7/2)+ \sqrt{133/4}\)
x<-9.266 OR x > 2.266

SUFFICIENT

Answer D

I don't know why people are going for A .

Also How x^2+7x−21>0
gives x<-7 U x >3. This is a wrong factorization. So either question need s to be changed or Answer needs to be changed... Bunuel
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answer is D. It can be drawn from statement 2 that x<-7 and x>3 thus we now x is not equal to 3.
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shashankism
Bunuel
Is x = 3?

(1) \(x^2 + 6x + 9 = 0\)

(2) \(x^2 + 7x - 21 > 0\)


Given : nothing
DS: x=3

Statement 1 : (1) \(x^2 + 6x + 9 = 0\)
\((x+3)^2 = 0\)
x= -3
SUFFICIENT

Statement 2 :

\(x^2 + 7x - 21 > 0\)

\(x^2 + 2*(7/2)x + 49/4 - 49/4 - 21 > 0\)
\((x+7/2)^2 - 133/4 > 0\)
\((x+7/2)< - \sqrt{133/4} or (x+7/2)> \sqrt{133/4}\)
\(x<-7/2- \sqrt{133/4} or (x>-7/2)+ \sqrt{133/4}\)
x<-9.266 OR x > 2.266

SUFFICIENT

Answer D

I don't know why people are going for A .

Also How x^2+7x−21>0
gives x<-7 U x >3. This is a wrong factorization. So either question need s to be changed or Answer needs to be changed... Bunuel
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