Solve the equation #(2x-3)(2x-1)(2x+1)(2x+3)=3465# ?
1 Answer
The solutions are
Explanation:
We start by multiplying out.
We can do this easily by recognizing that
#(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
#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!