Adding infinite square roots

**BG5965** · June 12th 2009, 07:44 AM

Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!

Any help is appreciated.

Thanks in advance.

BG

**alexmahone** · June 12th 2009, 07:51 AM

Let the expression be s.

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

Rejecting the negative root,

$s=3$

**TheEmptySet** · June 12th 2009, 07:54 AM

Originally Posted by BG5965

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

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

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

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

so you get

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

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

**BG5965** · September 1st 2009, 01:40 AM

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

Originally Posted by TheEmptySet

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

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

Can someone please explain that further? Thanks BG

**mr fantastic** · September 1st 2009, 02:14 AM

Originally Posted by BG5965

Hi, I have a 'different' (not necessarily difficult) algebra problem here. Please help!

Any help is appreciated.

Thanks in advance.

BG

This looks like part of an IB portfolio question. You can discuss the matter with me via pm.

Thread closed.

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