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

Question

Theodore Amidon · Accepted Answer

Use Cardano's method to find real root:

#x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))#

and related Complex roots.

First subtract #2# from both sides to get this equation into standard form: #x^3+2x^2-2 = 0# #color(white)()# Discriminant The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula: #Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd# In our example, #a=1#, #b=2#, #c=0# and #d=-2#, so we find: #Delta = 0+0+64-108+0 = -44# Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another. #color(white)()# Tschirnhaus transformation To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation. #0=27(x^3+2x^2-2)# #color(white)(0)=27x^3+54x^2-54# #color(white)(0)=(3x+2)^3-12(3x+2)-38# #color(white)(0)=t^3-12t-38# where #t=(3x+2)# #color(white)()# Cardano's method We want to solve: #t^3-12t-38=0# Let #t=u+v#. Then: #u^3+v^3+3(uv-4)(u+v)-38=0# Add the constraint #v=4/u# to eliminate the #(u+v)# term and get: #u^3+64/u^3-38=0# Multiply through by #u^3# and rearrange slightly to get: #(u^3)^2-38(u^3)+64=0# Use the quadratic formula to find: #u^3=(38+-sqrt((-38)^2-4(1)(64)))/(2*1)# #=(38+-sqrt(1444-256))/2# #=(38+-sqrt(1188))/2# #=(38+-6sqrt(33))/2# #=19+-3sqrt(33)# Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root: #t_1=root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33))# and related Complex roots: #t_2=omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33))# #t_3=omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33))# where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#. Now #x=1/3(-2+t)#. So the roots of our original cubic are: #x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))# #x_2 = 1/3(-2+omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33)))# #x_3 = 1/3(-2+omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33)))#

Serenity Johnson · Answer

undefined To solve the equation $x^3 + 2x^2 = 2$, you can follow these steps:

1. Subtract 2 from both sides to isolate the polynomial:
$$x^3 + 2x^2 - 2 = 0.$$

2. Try to factor the polynomial. In this case, you can't factor it easily, so you may need to use other methods.

3. One method is to use numerical methods or graphing technology to approximate the roots.

4. Another method is to use the Rational Root Theorem, which states that any rational roots of the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0$ must be of the form ±p/q, where p is a factor of the constant term $a_0$ and q is a factor of the leading coefficient $a_n$.

5. By trying various factors of 2 (the constant term) and 1 (the leading coefficient), you can find the rational roots of the polynomial.

6. Once you've found a root, you can use polynomial division or synthetic division to divide the polynomial by $x - \text{root}$ to obtain a quadratic equation.

7. Solve the quadratic equation to find the remaining roots, which may be real or complex.

8. Thus, you find all the roots of the original equation.