Question

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

Answer 1

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!