What is the exact value of #sqrt(10+sqrt(10+sqrt(10+...)))# ?

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Answer 1 · 2016-09-27T03:33:57

Let #x = sqrt(10 + sqrt(10 + sqrt(10 + ...)))#

Then #x = sqrt(10 + x)#

Solving this equation for x:

#x^2 = (sqrt(10 + x))^2#

#x^2 = 10 +x#

#x^2 - x - 10 = 0#

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-(-1) +- sqrt(-1^2 - 4 xx 1 xx -10))/(2 xx 1)#

#x = (1 +- sqrt(41))/2#

However, #x = (1- sqrt(41))/2# is extraneous since it doesn't satisfy the original equation.

Hence, the value of the expression #sqrt(10 + sqrt(10 + sqrt(10 + ...)))# is #(1 + sqrt(41))/2#.

Hopefully this helps!