We have equation: #a(x+3)^3=(x+3)e^(abs(x+1));ainRR#.How many real roots has this equation?

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Answer 1 · 2017-06-03T17:23:02

See below.

Making #b = e^abs(x+1) > 0# and #xi = x+3# we have

#a xi ^3-b xi = 0# or #(a xi^2-b)xi = 0# Here the roots are

#xi = 0 rArr x = -3# and

#a xi^2-b=0 rArr xi^2 = b/a# for #a ne 0# and then if #a > 0#

#xi = x+3 = pm e^((abs(x+1))/2)/sqrt a# and now using the so called Lambert function #W(cdot)#

we have

#x = -3-2W(pm1/(2 sqrt(a) e))#