1. ## Adding infinite square roots

Any help is appreciated.

BG

2. Let the expression be s.

$\displaystyle s^2=6+s$
$\displaystyle s^2-s-6=0$
$\displaystyle (s-3)(s+2)=0$
s=3 or -2

Rejecting the negative root,

$\displaystyle s=3$

3. Originally Posted by BG5965

Any help is appreciated.

BG
Well to simplify things lets suppose that the sum does exist, then

$\displaystyle x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}$

Then notice that red part is the "same" as x

$\displaystyle x=\sqrt{6+{\color{red} \sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}}$

so you get

$\displaystyle x=\sqrt{6+x} \iff x^2=6+x \iff x^2-x-6=0 \iff (x-3)(x+2)=0$

We reject the negative solution because we are using only the posative square root. so $\displaystyle x=3$

4. Hi, the only thing I don't understand is:

Originally Posted by TheEmptySet
Then notice that red part is the "same" as x

$\displaystyle x=\sqrt{6+{\color{red} \sqrt{6+\sqrt{6+\sqrt{6}+ ...}}}}$
Can someone please explain that further? Thanks BG

5. Originally Posted by BG5965