How do I solve this equation ?

#3x^4-5x^3+2=0#

2 Answers
MattyMatty
Feb 24, 2018

#"See explanation"#

Explanation:

#"First apply the rational roots theorem to find rational roots."#
#"We find "x=1" as rational root."#
#"So "(x-1)" is a factor. We divide that factor away : "#

#3 x^4 - 5 x^3 + 2 = (x-1)(3x^3-2x^2-2x-2)#

#"We have a remaining cubic equation that has no rational roots."#
#"We can solve it with the substitution of Vieta method."#

#x^3 - (2/3) x^2 - (2/3) x - 2/3 = 0#

#"Substitute "x = y+2/9". Then we get"#

#y^3 - (22/27) y - (610/729) = 0#

#"Substitute "y = (sqrt(22)/9) z". Then we get"#

#z^3 - 3 z - 5.91147441 = 0#

#"Substitute "z = t + 1/t". Then we get"#

#t^3 + 1/t^3 - 5.91147441 = 0#

#"Substituting "u = t^3", yields the quadratic equation :"#

#u^2 - 5.91147441 u + 1 = 0#

#"A root of this quadratic equation is u=5.73717252."#
#"Substituting the variables back, yields :"#

#t = root(3) (u) = 1.79019073#
#z = 2.34879043.#
#y = 1.22408929.#
#x = 1.44631151.#

#"The other roots are complex :"#
#-0.38982242 pm 0.55586071 i.#
#"(They can be found by dividing away "(x-1.44631151))#

George C.
Feb 24, 2018

The rational real zero is #x=1#.

Then there is an irrational real zero:

#x_1 = 1/9(2+root(3)(305+27sqrt(113))+root(3)(305-27sqrt(113)))#

and related non-real complex zeros.

Explanation:

Given:

#3x^4-5x^3+2 = 0#

Note that the sum of the coefficients is #0#.

That is: #3-5+2 = 0#

Hence we can deduce that #x=1# is a zero and #(x-1)# a factor:

#0 = 3x^4-5x^3+2#

#color(white)(0) = (x-1)(3x^3-2x^2-2x-2)#

The remaining cubic is somewhat more complicated...

Given:

#f(x) = 3x^3-2x^2-2x-2#

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=3#, #b=-2#, #c=-2# and #d=-2#, so we find:

#Delta = 16+96-64-972-432 = -1356#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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=243f(x)=729x^3-486x^2-486x-486#

#=(9x-2)^3-66(9x-2)-610#

#=t^3-66t-610#

where #t=(9x-2)#

Cardano's method

We want to solve:

#t^3-66t-610=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-22)(u+v)-610=0#

Add the constraint #v=22/u# to eliminate the #(u+v)# term and get:

#u^3+10648/u^3-610=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2-610(u^3)+10648=0#

Use the quadratic formula to find:

#u^3=(610+-sqrt((-610)^2-4(1)(10648)))/(2*1)#

#=(610+-sqrt(372100-42592))/2#

#=(610+-sqrt(329508))/2#

#=(610+-54sqrt(113))/2#

#=305+-27sqrt(113)#

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)(305+27sqrt(113))+root(3)(305-27sqrt(113))#

and related Complex roots:

#t_2=omega root(3)(305+27sqrt(113))+omega^2 root(3)(305-27sqrt(113))#

#t_3=omega^2 root(3)(305+27sqrt(113))+omega root(3)(305-27sqrt(113))#

where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Now #x=1/9(2+t)#. So the roots of our original cubic are:

#x_1 = 1/9(2+root(3)(305+27sqrt(113))+root(3)(305-27sqrt(113)))#

#x_2 = 1/9(2+omega root(3)(305+27sqrt(113))+omega^2 root(3)(305-27sqrt(113)))#

#x_3 = 1/9(2+omega^2 root(3)(305+27sqrt(113))+omega root(3)(305-27sqrt(113)))#

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