Quantitative Aptitude Ques 1499
Question: $ \sqrt{56+\sqrt{56+\sqrt{56}}}+…\div 2^{2}=? $
Options:
A) 0
B) 1
C) 2
D) 8
E) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ \sqrt{56+\sqrt{56+\sqrt{56}}}+…=x $ $ \sqrt{56+x}=x $ On squaring both side, we get $ {{(\sqrt{56+x})}^{2}}=x^{2} $
$ \Rightarrow $ $ 56+x=x^{2} $
$ \Rightarrow $ $ x^{2}-x-56=0 $
$ \Rightarrow $ $ x^{2}-8x+7x-56=0 $
$ \Rightarrow $ $ x,(x-8)+7(x-8)=0 $
$ \Rightarrow $ $ x=8 $
$ \therefore $ $ \sqrt{56+\sqrt{56+\sqrt{56}}}=8 $ So, $ \frac{8}{{{(2)}^{2}}}=\frac{8}{4}=2 $
BETA
