Complex Numbers And Quadratic Equations question 165
Question: If $ x=\sqrt{1+\sqrt{1+\sqrt{1+…….\text{to infinity}}}}, $ then x =
Options:
A) $ \frac{1+\sqrt{5}}{2} $
B) $ \frac{1-\sqrt{5}}{2} $
C) $ \frac{1\pm \sqrt{5}}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ x=\sqrt{1+\sqrt{1+\sqrt{1+…..}}} $ to $ \infty $ $ \infty $ We have $ x=\sqrt{1+x} $
Þ $ x^{2}=1+x\Rightarrow x^{2}-x-1=0 $
Þ $ x=\frac{1\pm \sqrt{1+4}}{2}=\frac{1\pm \sqrt{5}}{2} $ As $ x>0 $ , we get $ x=\frac{1+\sqrt{5}}{2} $
BETA
