Bunuel wrote:

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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