Nested Radicals........

Calculus Level 1

6 + 6 + 6 + … 3 3 3 { \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\ldots}}} }

Find value of this expression

1 2 4 3

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Caleb Townsend
Mar 1, 2015

x = 6 + x 3 x = \sqrt[3]{6 + x} x 3 = 6 + x x^3 = 6 + x x 3 − x − 6 = 0 x^3 - x - 6 = 0 ( x − 2 ) ( x 2 + 2 x + 3 ) = 0 (x-2)(x^2 + 2x + 3) = 0 The discriminant of x 2 + 2 x + 3 x^2 + 2x + 3 is negative, thus the only real root of the equation above is x = 2 x = \boxed{2}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Hello Caleb! How did you factor the cubic equation? Is there a method for doing this that I don't know of, or was it trial and error? I graphed it...

Dylan Scupin-Dursema - 6 years, 3 months ago

Log in to reply

Use the factor theorem to obtain (x-2) as a factor, then use an inspection method to obtain the second bracket

Curtis Clement - 6 years, 3 months ago

image image
This generalizes as

image image

Now putting x=2 , we get the answer.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...