Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!
Any help is appreciated.
Thanks in advance.
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 $\displaystyle x=3$