What is #sqrt(12+sqrt(12+sqrt(12+...)))# ?

1 Answer
George C.
Feb 2, 2017

#sqrt(12+sqrt(12+sqrt(12+...))) = 4#

Explanation:

Suppose:

#color(blue)(c = sqrt(12+sqrt(12+sqrt(12+...))))#

Note that #c >= 0# by the definition of #sqrt(...)#

Then:

#color(green)(sqrt(12+color(blue)(c)) = sqrt(12+color(blue)(sqrt(12+sqrt(12+sqrt(12+...)))))) = color(blue)(c)#

Squaring both ends, we find:

#12+c = c^2#

Note that squaring both sides of an equation results in an equation which must hold in order that the original equation holds, but is not necessarily sufficient. In this example, we will find a spurious negative solution for #c#, which we have already noted is excluded.

Subtract #12+c# from both sides to find:

#0 = c^2-c-12 = (c-4)(c+3)#

So #c = 4# or #c = -3#

We can discard the extraneous value #c = -3# which was introduced when we squared both ends of the equation.

So the correct solution is #c=4#.

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