How do you solve #x^3 + 4x^2 - x# using the quadratic formula?

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Answer 1 · 2015-10-25T10:22:30

#x={0,-2+-sqrt5}#

#x^3+4x^2-x=0# is a cubic equation. You first have to factor #x# out.

#[1]" "x^3+4x^2-x=0#

#[2]" "x(x^2+4x-1)=0#

The first root is #x=0# (from the #x# you factored out). You can use the quadratic formula to find the other two roots from #x^2+4x-1#.

#a=1#
#b=4#
#c=-1#

#[3]" "x=[-b+-sqrt(b^2-4ac)]/(2a)#

#[4]" "x=[-4+-sqrt(4^2-4(1)(-1))]/(2(1))#

#[5]" "x=[-4+-sqrt(20)]/2#

#[6]" "x=[-4+-2sqrt(5)]/2#

#[7]" "color(blue)(x=-2+-sqrt5)#

So the roots of the equation are:

#x={0,-2+-sqrt5}#