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

Question

Eli Assouline · Accepted 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.

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#.

Eva Abramson · Answer

undefined To solve the equation y = x^3 + 3x^2 - 6x using the quadratic formula, we first rearrange the equation to set it equal to zero:

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

Then, we apply the quadratic formula, where a = 1, b = 3, and c = -6:

x = [-b ± √(b^2 - 4ac)] / (2a)

Substitute the values of a, b, and c into the formula:

x = [-(3) ± √((3)^2 - 4(1)(-6))] / (2 * 1)

Calculate inside the square root:

x = [-(3) ± √(9 + 24)] / 2

x = [-(3) ± √(33)] / 2

Therefore, the solutions are:

x = (-3 ± √33) / 2