How do you find the solutions to the equation #x^4-6x^3+5x^2+30x-50=0#?

1 Answer
Aug 14, 2016

Answer:

This quartic has roots: #3+-i# and #+-sqrt(5)#

Explanation:

#f(x) = x^4-6x^3+5x^2+30x-50#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-50# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-5, +-10, +-25, +-50#

None of these work. So #f(x)# has no rational zeros.

#color(white)()#
Tschirnhaus transformation

First make a linear substitution called a Tschirnhaus transformation in order to simplify the quartic:

#16f(x) = 16x^4-96x^3+80x^2+480x-800#

#=(2x-3)^4-34(2x-3)^2+144(2x-3)-143#

#=t^4-34t^2+144t-143#

where #t = (2x-3)#

Since this quartic in #t# is monic and has no #t^3# term, it can be expressed as the product of two monic quadratics with opposite middle terms:

#t^4-34t^2+144t-143#

#=(t^2-ax+b)(t^2+ax+c)#

#=t^4+(b+c-a^2)t^2+(b-c)at+bc#

Equating coefficients and rearranging a little, we get:

#{ (b+c = a^2-34), (b-c = 144/a), (bc = -143) :}#

So we find:

#(a^2-34)^2 = (b+c)^2 = (b-c)^2+4bc = 20736/a^2-572#

Expanding, this becomes:

#(a^2)^2-68(a^2)+1156 = 20736/((a^2))-572#

Multiply through by #(a^2)# and rearrange a little to get:

#(a^2)^3-68(a^2)^2+1728(a^2)-20736=0#

By the rational roots theorem, the possible rational roots of this cubic are any factors of #20736 = 2^8*3^4#

That's rather a lot to try, but there are various methods to speed up the search and eventually find #a^2=36# is a zero.

Let #a=6#

Then:

#b = 1/2(a^2-34+144/a) = 1/2(36-34+24) = 13#

#c = 1/2(a^2-34-144/a) = 1/2(36-34-24) = -11#

Hmmm. We could have spotted that one fairly easily from the simultaneous equations by looking at possible factorings of #bc =-143#.

Well however we get there we now have two quadratics to solve:

#0 = t^2-6t+13 = (t-3)^2 + 2^2 = (t-3-2i)(t-3+2i)#

#0 = t^2+6t-11 = (t+3)^2-(2sqrt(5))^2 = (t+3-2sqrt(5))(t+3+2sqrt(5))#

Then #x=1/2(t+3)#

So we get solutions to our original quartic:

#x=1/2((3+-2i)+3)=3+-i#

#x=1/2((-3+-2sqrt(5))+3) = +-sqrt(5)#