How do you solve #x^3 + 4x^2 - x=0#?

1 Answer
Gió
Jul 15, 2015

I found three solutions as follows:
#x_1=0#
#x_2=-2+sqrt(5)#
#x_3=-2-sqrt(5)#

Explanation:

You can start collecting #x# to get:
#x(x^2+4x-1)=0#
So you get:
#x_1=0#
#x^2+4x-1=0# that can be solved using the Quadratic Formula as:
#x_(2,3)=(-4+-sqrt(16+4))/2# which gives you two additional solutions:
#x_2=(-4+sqrt(20))/2=-2+sqrt(5)#
and:
#x_3=(-4-sqrt(20))/2=-2-sqrt(5)#

Where I used the fact that #sqrt(20)=sqrt(5*4)=2sqrt(5)#

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