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        riddles >> easy >> Nested radical equations
        
(Message started by: NickH on Jul 27^th, 2004, 2:01pm)

Title: Nested radical equations
Post by NickH on Jul 27^th, 2004, 2:01pm

Let x be a real number. Find x if:

1) x = [sqrt](4 + [sqrt](4 + x))
2) x = [sqrt](4 + [sqrt](4 - x))
3) x = [sqrt](4 - [sqrt](4 + x))
4) x = [sqrt](4 - [sqrt](4 - x))

(Four questions.)

Title: Re: Nested radical equations
Post by Grimbal on Jul 28^th, 2004, 6:05am

::[hide]
1) x = sqrt(4+x)
x² - x - 4 = 0 and x>=0
x₁ = (1 + sqrt(17))/2

2) x = sqrt(4+sqrt(4-x))
note: x must be in [2,4] because 4-x>0 and x=sqrt(4+...)
x² = 4+sqrt(4-x)
x²-4 = sqrt(4-x)
x⁴-8x²+16 = 4-x
x⁴-8x²+x+12 = 0
(x²-x-3)(x²+x-4) = 0
x = (1 [pm] sqrt(13))/2 or x = (-1 [pm] sqrt(17))/2
The only valid solution to is x₂ = (1+sqrt(13))/2
The other solution work if you compute negative roots.

3) x = sqrt(4-sqrt(4+x))
the solution is x3 is such that
x₂ = sqrt(4+x₃) or x₃ = x₂²-4
((1+sqrt(13))/2)²-4 = (14+2sqrt(13)-16)/4 = (-1+sqrt(13))/2
x₃ = (-1+sqrt(13))/2

4) Just like 1)
x = sqrt(4-x)
x² + x - 4 = 0 and 0<=x<=4
x₄ = (-1 + sqrt(17))/2
::[/hide]

Title: Re: Nested radical equations
Post by NickH on Jul 29^th, 2004, 3:05pm

::[hide]For 1, how do you know that x = sqrt(4+x) yields the only solution to the original equation? Similarly for 4.[/hide]::

Title: Re: Nested radical equations
Post by Grimbal on Jul 29^th, 2004, 4:08pm

Hm.. you are right. The shortcut was not very correct.
::[hide]
But if it works the way it worked for 2 and 3, there should be 4 solutions, but the 4 solutions solve the variations where you choose the negative value of the one or the other square root. So only one should work.
But now that you mention it, with chaos theory in the back of my head, I suspect it might not work for other values of 4.

Anyway, nobody came up with another solution, so it should be the only one. ;D
[/hide]::