If $x^4 + x^{-4} = 47, x > 0, $ then the value of $(2x - 3)^2$ is :

5

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

Given,

If $x^4 + x^{-4} = 47, x > 0, $

then x² + \(\frac{1}{x^2}\) = \(\sqrt {47 + 2}\) = 7

and x + \(\frac{1}{x}\) = \(\sqrt {7 + 2}\) = 3

we can write is as = x² + 1 = 3x or x² - 3x = -1

also, 4x² - 12x = -4

then the value of $(2x - 3)^2$ is = 4x² + 9 - 12x

= $(2x - 3)^2$ is =  9 - 4 = 5