Complex Numbers And Quadratic Equations question 802
Question: If x is real and satisfies $ x+2>\sqrt{x+4}, $ then [AMU 1999]
Options:
A) $ x<-2 $
B) $ x>0 $
C) $ -3<x<0 $
D) $ -3<x<4 $
Show Answer
Answer:
Correct Answer: B
Solution:
Given, $ x+2>\sqrt{x+4} $
$ \Rightarrow {{(x+2)}^{2}}>(x+4) $
$ \Rightarrow x^{2}+4x+4>x+4 $
$ \Rightarrow x^{2}+3x>0 $
$ \Rightarrow x(x+3)>0 $
Þ x < - 3 or x > 0
$ \Rightarrow x>0 $ .
BETA
