What are the roots of #38r^3-27r^2-27r-27 = 0# ?

1 Answer
Feb 5, 2017

Answer:

Real root:

#x = 3/2#

Complex roots:

#x = -15/38+-(3sqrt(51))/38i#

Explanation:

#f(x) = 38r^3-27r^2-27r-27#

By the rational roots 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 #-27# and #q# a divisor of the coefficient #38# of the leading term.

That means that the only possible rational zeros are:

#+-1/38, +-1/19, +-3/38, +-3/19, +-9/38, +-1/2, +-27/38, +-1, +-27/19, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27#

That's rather a lot of possibilities to try, so let's see how we can narrow down the search...

First note that the pattern of signs of the coefficients of #f(x)# is #+ - - -#. By Descartes' Rule of Signs, since this has one change, we can deduce that #f(x)# has exactly one positive real zero.

We can observe that:

#f(1) = 38-27-27-27 = -43 < 0#

So the positive Real zero is greater than #1#.

We also find:

#f(3) = 38(27)-27(9)-27(3)-27 = 27(38-9-3-1) = 27*25#

#= 675 > 0#

So let's try:

#f(3/2) = 38(27/8)-27(9/4)-27(3/2)-27 = 27(38/8-9/4-3/2-1)#

#=27(19-9-6-4)/4 = 0#

So #x=3/2# is a zero and #(2x-3)# a factor:

#38r^3-27r^2-27r-27 = (2x-3)(19x^2+15x+9)#

We can find the zeros of the remaining quadratic using the quadratic formula with #a=19#, #b=15#, #c=9#...

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-15+-sqrt(15^2-4(19)(9)))/(2*19)#

#color(white)(x) = (-15+-sqrt(225-684))/38#

#color(white)(x) = (-15+-3sqrt(51)i)/38#

#color(white)(x) = -15/38+-(3sqrt(51))/38i#