How do you solve #(x^2 - 16)(x - 3)^2 + 9x^2 = 0#?

1 Answer
Sep 5, 2015

Answer:

Find:

#(x^2-16)(x-3)^2+9x^2 = (x^2+2x-6)(x^2-8x+24)#

hence Real roots #x = -1 +-sqrt(7)#

Explanation:

Let #f(x) = (x^2-16)(x-3)^2+9x^2#

#= (x^2-16)(x^2-6x+9)+9x^2#

#= x^4-6x^3+2x^2+96x-144#

If #abs(x) >= 4# then #(x^2-16) >= 0#, #(x-3)^2 > 0# and #9x^2 > 0#, so #f(x) > 0#.

So the only Real roots of #f(x) = 0# are in the range #-4 < x < 4#.

The rational root theorem tells us that any rational roots of #f(x) = 0# must be of the form #p/q# in lowest terms, where #p, q in ZZ#, #q > 0#, #p# is a divisor of the constant term #144# and #q# is a divisor of the coefficient #1# of the leading term.

So the only possible rational roots are integer divisors of #144# strictly between #-4# and #4#. That gives the only possibilities as #+-1#, #+-2#, #+-3#. None of these work, so #f(x) = 0# has no rational roots.

Next, suppose #f(x) = (x^2+ax+b)(x^2+cx+d)# for some integers #a#, #b#, #c# and #d#.

What can we find out about these integers?

#(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd#

Equating coefficients, we get:

(i) #a+c=-6#
(ii) #b+d+ac=2#
(iii) #ad+bc=96#
(iv) #bd=-144#

Suppose #a# is odd.
Then from (i) #c# is also odd, so #ac# is odd.
Then from (ii) #b+d# is odd, so one of #b# and #d# is odd and the other even.
Then one of #ad# and #bc# is odd and the other even.
So #ad+bc# is odd.
But (iii) tells us that #ad+bc=96# which is even, so we have a contradiction.
Therefore #a# is even.

Since #a# is even, then from (i) we get #c# is even too. Thus #ac# is a multiple of #4#.
Then from (ii), we find #b+d# is even but not a multiple of #4#.
From (iv), at least one of #b# and #d# is even, hence they both are.

That leaves us with the following possibilities for #(b, d)# with #b < d#:

#(-72, 2)#, #(-24, 6)#, #(-18, 8)#, #(-8, 18)#, #(-6, 24)#, #(-2, 72)#

..with sums:

#-70#, #-18#, #-10#, #10#, #18#, #70#

..which from (ii) give us possible values for #ac#:

#72#, #20#, #12#, #-8#, #-16#, #-68#

..which in combination with (i) give us possible integer values of #(a, c)#:

(none), (none), (none), (none), #(-8, 2)# / #(2, -8)#, (none)

So #(b, d) = (-6, 24)# and #(a, c) = (-8, 2) " or " (2, -8)#

From (iii) we quickly find:

#(a, c) = (2, -8)#

Hence:

#f(x) = (x^2+2x-6)(x^2-8x+24)#

Using the quadratic formula, the first of these factors has roots:

#x = -1 +-sqrt(7)#

and the second only has complex roots.