How do you solve #1/((x - 1)(x - 2)) + 1/((x - 2)(x - 3) )+ 1/((x - 3)(x - 4)) = 1/6#?

1 Answer
May 8, 2016

Answer:

#x=-2# or #x=7#

Explanation:

Given:

#1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4)) = 1/6#

Multiply through by #6(x-1)(x-2)(x-3)(x-4)# to get:

#6(x-3)(x-4)+6(x-1)(x-4)+6(x-1)(x-2) = (x-1)(x-2)(x-3)(x-4)#

Multiply out:

#6(x^2-7x+12)+6(x^2-5x+4)+6(x^2-3x+2) = x^4-10x^3+35x^2-50x+24#

#rarr#

#18x^2-90x+108 = x^4-10x^3+35x^2-50x+24#

Subtract the left hand side from the right to get:

#x^4-10x^3+17x^2+40x-84 = 0#

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-84# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1#, #+-2#, #+-3#, #+-4#, #+-6#, #+-7#, #+-12#, #+-14#, #+-21#, #+-28#, #+-42#, #+-84#

Substituting #x=2# into the quartic we find:

#x^4-10x^3+17x^2+40x-84 = 16-80+68+80-84 = 0#

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

#x^4-10x^3+17x^2+40x-84 = (x-2)(x^3-8x^2+x+42)#

Substituting #-2# into this cubic we find:

#x^3-8x^2+x+42 = -8-32-2+42 = 0#

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

#x^3-8x^2+x+42 = (x+2)(x^2-10x+21)#

To factor and find the zeros of the remaining quadratic, note that #3+7 = 10# and #3*7 = 21#, so:

#x^2-10x+21 = (x-3)(x-7)#

So the remaining zeros are #x=3# and #x=7#.

So all the zeros of our quartic polynomial are:

#-2, 2, 3, 7#

Note that the values #2# and #3# are not solutions of the original rational equation, since they result in zero denominators.

So the solutions of the original rational equation are #x=-2# and #x=7#