# problem solving - square roots

• Dec 4th 2009, 01:38 AM
problem solving - square roots
This problem was in my calculus exercise, yet I don't understand why:

Solve the equation :
$\displaystyle x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}$
(Show there are not other solutions to the ones you found)

Now, solving it regularly using simple algebra got me into an eighth-degree polynomial, which reminded my my teachers have just told us there are proofs for why there are no 'formulas' for solving polynomial equations higher than 5.

Anyway, how can I solve this problem using calculus tools? I thought of doing the following thing:
We know that $\displaystyle x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}$, therefore:
$\displaystyle x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}=\sqrt{5+\sqrt{5+\sq rt{5+\sqrt{5+\sqrt{5+\sqrt{5+x}}}}}}$.

Let the first expression be $\displaystyle a_3$, ane the second expression be $\displaystyle a_6$. Therefore:

$\displaystyle x=a_{3n}=\sqrt{5+\sqrt{5+\sqrt{5+...}}}$
Therefore,
$\displaystyle lim(x)=lim(a_{3n})$

but how do I find it :) ?
• Dec 4th 2009, 06:02 AM
Does anyone have any idea (Thinking)?

This looks like a nice one...
• Dec 4th 2009, 06:48 AM
Defunkt
Well, if you think about it you can write:

$\displaystyle x = \sqrt{5+x}$ (can you understand why?)
So all you need to do is solve the quadratic equation you get after squaring both sides.
• Dec 4th 2009, 06:56 AM
Quote:

Originally Posted by Defunkt
Well, if you think about it you can write:

$\displaystyle x = \sqrt{5+x}$ (can you understand why?)
So all you need to do is solve the quadratic equation you get after squaring both sides.

Hmm.. I can only write $\displaystyle x = \sqrt{5+x}$, then try to see if it works and get back again $\displaystyle x = \sqrt{5+x}$. I don't think this is how you got to it, and how I should understand it.

This means it is a solution, but how can I know it's the only one?
• Dec 4th 2009, 08:19 AM
Defunkt
$\displaystyle x = \sqrt{5+\sqrt{5+\sqrt{5+x}}} \Rightarrow x = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+x} }}}}}$ $\displaystyle \Rightarrow x = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+ \sqrt{...}}}}}}}$

So essentially, this is equivalent to $\displaystyle x= \sqrt{5+x}$, thus the solutions of the equation are the only ones which satisfy the condition for x.
• Dec 4th 2009, 09:22 AM