$\sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}$

By Krishnan Krishnan on Thursday, August 22, 2002 - 11:00 am:

What is

\sqrt{(6 + \sqrt{6 + \sqrt{6 + \dots}}}

By Dan Goodman on Thursday, August 22, 2002 - 11:31 am:

Here's a hint, suppose we call the answer

x

, so that

x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}

. What is

x^{2} - 6

? What is

x

By Krishnan Krishnan on Thursday, August 22, 2002 - 11:46 am:

x² -x-6=0 which leads to x=3. Is this okay? hearty thanks

By Chris Tynan on Thursday, August 22, 2002 - 06:56 pm:

Just to let you know, this is a past National Mathematics Competition question. The answer is as Krishnan said, 3, but what about the other root?

By Graeme Mcrae on Thursday, August 22, 2002 - 07:47 pm:

The other root is -2, which is extraneous. It was introduced when we started with

$x = \sqrt{6 + x}$
and then squared both sides of the equation to get

$x^{2} = 6 + x$

--Graeme

By Andre Rzym on Thursday, August 22, 2002 - 08:59 pm:

A more complex 'continued radical' type problem is discussed here if anyone is interested ...

Andre