#(x+1)(x+2)(x+3)(x+4)=1# Express this equation?

1 Answer
George C.
Feb 24, 2018

#x = -5/2+-sqrt(5/4+-sqrt(2))#

Explanation:

Given:

#(x+1)(x+2)(x+3)(x+4) = 1#

Note the symmetry about #x=5/2#

Let #t = x+5/2#

Then:

#1 = (t-3/2)(t-1/2)(t+1/2)(t+3/2)#

#color(white)(1) = (t^2-9/4)(t^2-1/4)#

#color(white)(1) = t^4-5/2t^2+9/16#

Subtract #1# from both ends to get:

#0 = t^4-5/2t-7/16#

Multiply both sides by #16# to get:

#0 = 16t^4-40t^2-7#

#color(white)(0) = (4t^2)^2-2(4t^2)(5)+25-32#

#color(white)(0) = (4t^2-5)^2-(4sqrt(2))^2#

#color(white)(0) = ((4t^2-5)-4sqrt(2))((4t^2-5)+4sqrt(2))#

#color(white)(0) = (4t^2-5-4sqrt(2))(4t^2-5+4sqrt(2))#

#color(white)(0) = 16(t^2-5/4-sqrt(2))(t^2-5/4+sqrt(2))#

Hence:

#t^2 = 5/4+-sqrt(2)#

That is:

#(x+5/2)^2 = 5/4+-sqrt(2)#

So:

#x = -5/2+-sqrt(5/4+-sqrt(2))#

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