How do you solve #x^ { 5} + 3x ^ { 4} + 16x ^ { 3} + 48x ^ { 2} + 63x + 189= 0#?

1 Answer
Tamir E.
Nov 14, 2017

For #x in R# - Real Numbers, regular
#x=-3#

--
(For #x in C# - just if one needs, not for precalculus
#x_1=-3#
#x_(2,3)=+-sqrt(7)i#
#x_(4,5)=+-3i#)

Explanation:

#x^5+3x^4+16x^3+48x^2+63x+189=0#
#=> (x^5+3x^4)+(16x^3+48x^2)+(63x+189)=0#
#=> x^4(x+3)+16x^2(x+3)+63(x+3)=0#

#(x+3)=k#
#=> kx^4+k16x^2+63k=0#
#=>k(x^4+16x^2+63)=0#

#(x+3)(x^4+16x^2+63)=0#

#x^2=t#
#=> (t^2+16x+63)=0#
#t_(1,2)=(-16+-sqrt(256-4*1*63))/(2*1)#
#=>...#
#t_1=(-16+2)/2=-7#
#t_2=(-16-2)/2=-9#
#=> (t+7)(t+9)=0#
#=>(x^2+7)(x^2+9)=0#

#(x+3)(x^2+7)(x^2+9)=0#
#x_1=-3#
#x_2=sqrt(-7)=NaN#
#x_3=sqrt(-9)=NaN#

#x=-3#

Complex (#x in C#):
#x_1=-3#
#x_(2,3)=sqrt(7)i#
#x_(4,5)=3i#

Related questions
Impact of this question
1335 views around the world
You can reuse this answer
Creative Commons License