Solve the equation #(2x-3)(2x-1)(2x+1)(2x+3)=3465# ?

1 Answer
Jun 26, 2017

The solutions are #x = +-4, x = +- 3sqrt(3/2)i#

Explanation:

We start by multiplying out.

We can do this easily by recognizing that #2x - 3 and 2x+ 3#, as well as #2x - 1 and 2x + 1# are differences of squares.

#(2x + 3)(2x- 3) = 4x^2 - 9#

#(2x + 1)(2x- 1) = 4x^2 - 1#

#(2x - 3)(2x - 1)(2x+ 1)(2x + 3) = (4x^2 - 9)(4x^2 - 1)#

#(2x- 3)(2x- 1)(2x+ 1)(2x+ 3) = 16x^4 - 36x^2 - 4x^2 + 9#

#(2x - 3)(2x- 1)(2x+ 1)(2x+ 3) = 16x^4 - 40x^2 + 9#

Therefore,

#16x^4 - 40x^2 + 9 = 3465#

It follows that

#16x^4 - 40x^2 - 3456 = 0#

#2x^4 - 5x^2 - 432 = 0#

We now let #y = x^2#.

#2y^2 - 5y - 432 = 0#

We can solve by factoring.

#2y^2 - 32y + 27y - 432 = 0#

#2y(y - 16) + 27(y - 16) = 0#

#(2y + 27)(y - 16) = 0#

#y = -27/2 and 16#

#x^2 = -27/2 and 16#

#x = +- 4 and +- 3sqrt(3/2) i#

Hopefully this helps!