Complex Numbers And Quadratic Equations question 749
Question: If $ x=\sqrt{6+\sqrt{6+\sqrt{6+….to\infty }},} $ then [Pb. CET 1999]
Options:
A) x is an irrational number
B) $ 2<x<3 $
C) $ x=3 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=\sqrt{6+x} $ , $ x>0\Rightarrow x^{2}=6+x,x>0 $
$ \Rightarrow x^{2}-x-6=0,x>0 $
$ \Rightarrow x=3,x>0 $ .
BETA
