# problem solving - square roots

• December 4th 2009, 02:38 AM
problem solving - square roots
This problem was in my calculus exercise, yet I don't understand why:

Solve the equation :
$x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}$
(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 $x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}$, therefore:
$x=\sqrt{5+\sqrt{5+\sqrt{5+x}}}=\sqrt{5+\sqrt{5+\sq rt{5+\sqrt{5+\sqrt{5+\sqrt{5+x}}}}}}$.

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

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

but how do I find it :) ?
• December 4th 2009, 07:02 AM
Does anyone have any idea (Thinking)?

This looks like a nice one...
• December 4th 2009, 07:48 AM
Defunkt
Well, if you think about it you can write:

$x = \sqrt{5+x}$ (can you understand why?)
So all you need to do is solve the quadratic equation you get after squaring both sides.
• December 4th 2009, 07:56 AM
Quote:

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

$x = \sqrt{5+x}$ (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 $x = \sqrt{5+x}$, then try to see if it works and get back again $x = \sqrt{5+x}$. 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?
• December 4th 2009, 09:19 AM
Defunkt
$x = \sqrt{5+\sqrt{5+\sqrt{5+x}}} \Rightarrow x = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+x} }}}}}
$
$\Rightarrow x = \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+ \sqrt{...}}}}}}}$

So essentially, this is equivalent to $x= \sqrt{5+x}$, thus the solutions of the equation are the only ones which satisfy the condition for x.
• December 4th 2009, 10:22 AM