The polynomial #8x^3-36x^2+22x+21=0# has roots which form an arithmetic progression. How do you find the roots?

2 Answers
MoominDave
Jun 13, 2018

Use the information that the roots are in arithmetic relation to reduce the order of the problem from cubic to quadratic

Explanation:

Firstly, let's write our equation with "monic" first term, i.e. with first coefficient equal to 1.
#8x^3-36x^2+22x-21=0#
#x^3-9/2x^2+11/4x-21/8=0#

We know from the fact that it has highest polynomial term cubic that there are three roots. We know from given info that the three roots are separated by the same number - write them as #a-b#, #a#, #a+b#, where #a# and #b# are to be determined (experience suggests writing them in this symmetric fashion is likely to ease algebraic manipulation at some point or other...). The extra bit of information given us (that the roots are in arithmetic progression) ought to allow us to reduce the equation from cubic to quadratic. So
#(x-(a-b))(x-a))(x-(a+b))=0#
Multiply out the two symmetric brackets:
#(x-a)(x^2-2ax+(a^2-b^2))=0#
(Note that the chosen symmetric way of writing the roots has made this a simpler expression that it would otherwise have been - it's always worth keeping your algebra as tidy as possible - less confusing, harder to make mistakes)

Multiply out the third bracket:
#x^3-3ax^2+(3a^2-b^2)x-a(a^2-b^2)=0#

Now we can immediately deduce #a#, the base of the arithmetic progression. In the second coefficient we have:
#-3a=-9/2#, so
#a=3/2#

In the third coefficient we have:
#3a^2-b^2=11/4#, so
#27/4-b^2=11/4#
#4-b^2=0#
#b=+-2#

So we have two solutions for #b#, but in fact these are the same solution - one is the other one backwards. So our three arithmetically progressing roots #a-b#, #a#, #a+b# are #-1/2#, #3/2#, #7/2#.

We now double-check our answer by substituting it back in to the original equation (this is important to do! I caught a mistake in my working by doing it...):
#(x+1/2)(x-3/2)(x-7/2)=0#
#(2x+1)(2x-3)(2x-7)=0#
#8x^3-36x^2+22x+21=0#, our original equation as hoped...

Steve M
Jun 13, 2018

#-1/2, 3/2, 7/2#

Explanation:

We have:

# 8x^3-36x^2+22x+21 = 0 #

We are given that the roots are in Arithmetic Progression. Let us denote the roots by #alpha#, #beta# and #gamma#, and let us denote the common difference of the AP by #d#, then we can write the three roots as:

# x_1 = alpha #
# x_2 = beta = alpha+d #
# x_2 = gamma = alpha+2d #

The sum of roots properties gives us:

# alpha + beta + gamma = -b/a #

# :. (alpha ) + (alpha + d) + (alpha+2d) = -(-36/8) #

# :. 3alpha +3d = 9/2 #

# :. alpha +d = 3/2 # ..... [A]

The product of roots properties gives us:

# alpha \ beta \ gamma = -d/a #

# :. (alpha)(alpha+d)(alpha+2d) = -21/8 #

# :. alpha(3/2)(alpha+2(3/2-alpha)) = -21/8 # (using [A])

# :. 3/2 alpha(alpha+3-2alpha) = -21/8 #

# :. 12 alpha(3-alpha) = -21 #

# :. 36 alpha-12alpha^2=-21 #

# :. 12alpha^2 -36alpha-21 = 0 #

And completing the square:

# alpha^2 -36/12alpha-21/12 = 0 #

# :. (alpha -3/2)^2-(3/2)^2-21/12 = 0 #

# :. (alpha -3/2)^2 = 21/12 +9/4 #

# :. alpha -3/2 = +-sqrt(4) #

# :. alpha = 3/2+-2 #

# :. alpha = -1/2, 7/2 #

And, with each of these solutions we get (using [A]):

Case (1):

# alpha = -1/2 => -1/2+d=3/2 => d = 2#

Leading to the roots:

# (alpha ),(alpha + d),(alpha+2d) # ie #-1/2, 3/2, 7/2#

Case (2):

# alpha = \ \ \ \ \ 7/2 => \ \ \ \ \ 7/2+d=3/2 => d = -2#

Leading to the roots:

# (alpha ),(alpha + d),(alpha+2d) # ie #7/2, 3/2, -1/2#

So, in both cases we have the roots:

#-1/2, 3/2, 7/2#

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