How do you solve #y=x^3 + 3x^2 - 6x # using the quadratic formula?

1 Answer
Jul 15, 2017

Answer:

This cubic has 3 roots: #x=0, -4.37# or #1.37#

We can solve it be dividing through by #x# and using the quadratic formula.

Explanation:

This is not a quadratic, it's a cubic, so it'll have 1-3 roots rather than 0-2 (think about the shape of a cubic compared to a parabola-shaped quadratic and how each can cut the x-axis).

It's a tricky one, but we can divide through by x and treat it as:

#y=x(x^2+3x-6)#

One root of this will be #x=0#: when #x=0#, then #y=0#, which is the definition of a root.

The other roots will be when the parenthesis is equal to #0#, so we have:

#x^2+3x-6=0#

And hey presto, we have a quadratic to solve! And we know how to do that: the quadratic formula. For a quadratic in the form:

#ax^2+bx+c=0#

We know:

#x=(-b+-sqrt(b^2-4ac))/(2a)=(-3+-sqrt(3^2-4xx1xx-6))/(2xx1)#
#=(-3+-sqrt(9+24))/(2)=(-3+-sqrt(33))/(2)=(-3+-sqrt(33))/(2)#
#=(-3-5.74)/2# or #(-3+5.74)/2#

Therefore #x=-4.37# or #1.37#

Remember that we had the other root, #x=0#, so the 3 roots all together are #x=0, -4.37# or #1.37#.