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

Given,  $$x^4 + x^{-4} = 47$$

$$=$$>  $$x^4+\frac{1}{x^4}+2=47+2$$

$$=$$>  $$\left(x^2+\frac{1}{x^2}\right)^2=49$$

$$=$$>   $$x^2+\frac{1}{x^2}=7$$

$$=$$>   $$x^2+\frac{1}{x^2}+2=7+2$$

$$=$$>   $$\left(x+\frac{1}{x}\right)^2=9$$

$$=$$>    $$x+\frac{1}{x}=3$$

$$=$$>   $$x^2-3x+1=0$$

$$=$$>   $$x^2-3x=-1$$

$$\therefore\ \left(2x-3\right)^2=4x^2+9-12x=4\left(x^2-3x\right)+9=4\left(-1\right)+9=5$$

Hence, the correct answer is Option A