How many complex roots does #6x^4+x^3-5=0# have?

2 Answers
Jul 15, 2015

#6x^4+x^3-5 = 0# has 2 complex roots
#color(white)("XXXX")#...but I hope someone has a better argument for this than what I've attempted here.

Explanation:

#6x^4+x^3-5 = (x+1)(6x^3-5x^2+5x-5)#
So
#color(white)("XXXX")##6x^4+x^3-5 = 0#
has, at least 1 Real root (#x=-1#)

The remaining factor equation: #6x^3-5x^2+5x-5 = 0#

  • has no (apparent) discontinuities
  • approaches #-oo# as #x# decreases
  • approaches #+oo# as #x# increases

and therefore most cross the X-axis at some point
#rArr# there is at least 1 more Real root
#color(white)("XXXX")##color(white)("XXXX")#graphing would seem to indicate a value between #(0,1)#
#color(white)("XXXX")##color(white)("XXXX")#but without attempting a messy factoring....

An polynomial equation with highest variable exponent of 4 may have 4 roots (some may disappear if two roots occur at the same value).

We have established that 2 of the roots are Real.

And the question is whether the remaining 2 are Real or Complex.

Taking the derivative of #6x^3-5x^2+5x-5# in order to establish points of inflection, gives us
#color(white)("XXXX")##18x^2-10x+5#
The discriminant of this (using the quadratic formula) indicates that there are no Real roots (#rarr# no Real points of inflection)
and since there are no points of inflection,
#color(white)("XXXX")##6x^3-5x^2+5x-5#
is a continuously increasing function which (therefore) only cross the X-axis at one point.

Therefore, there must be 2 Complex roots.

Jan 29, 2016

#2#

Explanation:

Factoring won't work very well, but the graph of a function is a great indicator of its types of roots.

Since this is a quartic equation (it has degree #4#), we know it will have #4# total roots.

Real roots occur when the graph of the function crosses the #x# axis.

The function's graph is:

graph{6x^4+x^3-5 [-4, 4, -6.33, 6.33]}

Here, we can see the two real roots at #x=-1# and #xapprox0.916#.

Since there are only two real roots, the other two must be complex roots. Note that this makes mathematical sense since complex roots will always come in pairs in polynomials with real coefficients such as this.