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!