How to solve this equation? #root(n)((x+1)^2)+root(n)((x-1)^2)=4root(n)(x^2-1)#

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Answer 1 · 2017-06-01T17:31:33

See below.

Calling #a = root(n)(x-1)# and #b = root(n)(x+1)# we have

#a^2+b^2=4ab# but

#(a-b)^2= a^2+b^2-2ab# so

#(a-b)^2=2ab# and then

#a-b=sqrt2 sqrt(ab)#

and also following the same procedure

#a+b=sqrt(6)sqrt(ab)#

solving now

#{(a+b=sqrt(6)sqrt(ab)),(a-b=sqrt2 sqrt(ab)):}#

we have

#b = (2+sqrt3)a# or

#b^n= (2+sqrt3)^n a^n# so

#x+1=(2+sqrt3)^n(x-1)# then

#x = ((2+sqrt3)^n+1)/((2+sqrt3)^n-1)#

but analogously

#x-1=(2+sqrt3)^n(x+1)# then

#x = -((2+sqrt3)^n+1)/((2+sqrt3)^n-1)#

and finally

#x = pm((2+sqrt3)^n+1)/((2+sqrt3)^n-1)#