#(sqrt(49+20sqrt6))^(sqrt(asqrt(asqrt(a...oo)#+#(5-2sqrt6)^(x^2+x-3 - sqrt(xsqrt(xsqrt(x...oo))))=10# where #a=x^2-3#, then x is?

Question

A) #-sqrt2#
B)#sqrt2#
C)#-2#
D)#2#

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Answer 1 · 2017-11-22T12:15:53

#x = 2#

Calling #sqrt[49 + 20 sqrt[6]] = 5 + 2 sqrt[6]= beta# we have

# (5 + 2 sqrt[6])^1+ (5- 2 sqrt[6])^1 = 10#

for

#sqrt(asqrt(asqrt(a...oo))) = 1# and

#x^2+x-3 - sqrt(xsqrt(xsqrt(x...oo))) = 1#

and such that

#a=x^2-3#

but

#sqrt(asqrt(asqrt(a...oo))) = a^(1/2+1/4+1/8+ cdots +1/2^k + cdots) = a^1 = 1#

and then

#1 = x^2-3 rArr x = 2#

then

#x^2+x-3 - sqrt(xsqrt(xsqrt(x...oo)))=1#

or

#1+2- sqrt(2sqrt(2sqrt(2...oo)))=1#

then #x = 2#