How do you find all the real and complex roots of #2x^4 + 3x^3 - x^2 + 5 = 0#?

1 Answer
Nov 20, 2016

Answer:

See explanation for a sketch of how to solve this algebraically...

Explanation:

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

This is an example of the worst possible case in quartic equations.

The full solution is too long, so I will just sketch it here...

  • Use a Tschirnhaus transformation to simplify the quartic to one with no cube term:

#2048f(x) = 4096x^4+6144x^3-2048x^2+10240#

#color(white)(2048f(x)) = (8x+3)^4-86(8x+3)^2+408(8x+3)+9709#

#color(white)(2048f(x)) = t^4-86t^2+408t+9709#

where #t = 8x+3#

  • Consider a factorisation of the form #(t^2-at+b)(t^2+at+c)# and equate coefficients to get:

#{ (b+c = a^2-86), (b-c = 408/a), (bc = 9709) :}#

  • Use #(b+c)^2 = (b-c)^2+4bc# to derive a cubic in #a^2#:

#(a^2)^3-172(a^2)^2-31440(a^2)-166464 = 0#

  • Use a Tschirnhaus transformation to simplify the cubic to one with no square term:

#27((a^2)^3-172(a^2)^2-31440(a^2)-166464)#

#= 27(a^2)^3-46644(a^2)^2-848880(a^2)-4494528#

#= (3a^2-172)^3-371712(3a^2-172)-63340544#

#= s^3-371712s-63340544#

where #s = 3a^2-172#

  • This cubic has #3# Real zeros, so use a trigonometric substitution of the form #s = k cos theta# with #k=704#

#0 = s^3-371712s-63340544#

#color(white)(0) = 704^3 cos^3 theta - 371712*704 cos theta - 63340544#

#color(white)(0) = 87228416(4 cos^3 theta - 3cos theta) - 63340544#

#color(white)(0) = 87228416 cos 3 theta - 63340544#

#color(white)(0) = 32768(2662 cos 3 theta - 1933)#

Hence:
#s = 704 cos(1/3cos^(-1)(1933/2662)+(2kpi)/3)# for #k=0,1,2#

The positive Real root occurs for #k=0#, so choose that to get:

#a = sqrt(1/3( 704 cos(1/3cos^(-1)(1933/2662))+172)#

  • Going back to our simultaneous equations in #a, b# and #c# we find:

#b = 1/2(a^2-86+408/a)#

#c = 1/2(a^2-86-408/a)#

  • Hence we have two quadratic equations to solve:

#t^2-at+1/2(a^2-86+408/a) = 0#

#t^2+at+1/2(a^2-86-408/a) = 0#

  • Solve these quadratic equations using the quadratic formula to find solutions to #t^4-86t^2+408t+9709=0#

  • Reverse the initial Tschirnhaus transformation using #x = 1/8(t-3)# to find the four (Complex) roots of the original quartic.