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