Question

$\sqrt{8+{\sqrt{8+{\sqrt{8+........}}}}}=x$ then x =?

• A. $$\frac{-1+\sqrt{33}}{2}$$
• B. $$\frac{7}{2}$$
• C. $$\frac{1+\sqrt{33}}{2}$$
• D. 3

Answer 1

Answer:

$$\frac{1+\sqrt{33}}{2}$$

$\because \sqrt{a+{\sqrt{a+{\sqrt{a+........}}}}}=x$ then${x}=\frac{1+\sqrt{4\times{a}+1}}{2}$

$\sqrt{8+{\sqrt{8+{\sqrt{8+........}}}}}=x$

$x=\frac{1+\sqrt{4\times{8}+1}}{2}$

$=\frac{1+\sqrt{33}}{2}$