Find the value of x

Find the value of x

\(x= \sqrt{20+\sqrt{20+\sqrt{20}}}\)

1. 20
2. 5
3. 2
4. 8

Question should be what is the approximate value of \(x\).

Obviously answer choice C (2) is out as \(\sqrt{20+some \ #}>4\).

Now, \(4<\sqrt{20}<5\):
\(x= \sqrt{20+\sqrt{20+\sqrt{20}}}= \sqrt{20+\sqrt{20+(# \ less \ than \ 5)}}= \sqrt{20+\sqrt{# \ less \ than \ 25}}= \sqrt{20+(# \ less \ than \ 5)}=\)

\(=\sqrt{# \ less \ than \ 25}=# \ less \ than \ 5\approx{5}\).

Answer: B.

Next, exactly 5 to be the correct answer question should be:

If the expression \(x=\sqrt{20+{\sqrt{20+\sqrt{20+\sqrt{20+...}}}}}\) extends to an infinite number of roots and converges to a positive number x, what is x?

\(x=\sqrt{20+{\sqrt{20+\sqrt{20+\sqrt{20+...}}}}}\) --> \(x=\sqrt{20+({\sqrt{20+\sqrt{20+\sqrt{20+...})}}}}\), as the expression under square root extends infinitely, then expression in brackets would equal to \(x\) itself so we can rewrite given expression as \(x=\sqrt{20+x}\). Square both sides \(x^2=20+x\) --> \(x=5\) or \(x=-4\). As given that \(x>0\) then only one solution is valid: \(x=5\).

Hope it helps.