The difference between the maximum real root and the minimum real root of the equation $$(x^2 - 5)^4 + (x^2 - 7)^4 = 16$$ is

The  given equation is $$(x^2 - 5)^4 + (x^2 - 7)^4 = 16$$

Let, $$x^2-5=a$$ then $$x^2-7=a-2$$

The equation, therefore, becomes, $$a^4+\left(a-2\right)^4=16$$. We simplify this equation to find real values of a , and then consequently find real values of x.

Now, $$a^4+\left(a-2\right)^4=16$$

or, $$\left(a^2\right)^2+\left(\left(a-2\right)^2\right)^2=16$$

or, $$\left(a^2-\left(a-2\right)^2\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$\left(\left(a+a-2\right)\times\ \left(a-a+2\right)\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$\left(2\times\ 2\left(a-1\right)\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$16\times\ \left(a-1\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$8\times\ \left(a-1\right)^2+\ a^2\times\ \left(a-2\right)^2=8$$

or, $$a^2\times\ \left(a-2\right)^2=8-\ 8\times\ \left(a-1\right)^2=8\left\{1-\left(a-1\right)^2\right\}=8\left(1-a+1\right)\left(1+a-1\right)=8a\left(2-a\right)$$

$$\therefore\ a^2\times\ \left(a-2\right)^2=8a\left(2-a\right)$$

or, $$a^2\times\ \left(a-2\right)^2-8a\left(2-a\right)=0$$

or, $$a^2\times\ \left(a-2\right)^2+8a\left(a-2\right)=0$$

or, $$a\times\ \left(a-2\right)\left\{a\left(a-2\right)+8\right\}=0$$

$$\therefore\ a=0$$ or $$a-2=0$$ or $$a\left(a-2\right)+8=0$$

From the first eqn, a=0, $$x^2-5=0\ =>x=\pm\ \sqrt{\ 5}$$

From the second eqn, a-2=0,  $$x^2-7=0\ =>x=\pm\ \sqrt{\ 7}$$

ANd from the third eqn, $$a\left(a-2\right)+8=0$$ $$a\left(a-2\right)+8=0\ =>\ a^2-2a+8=0$$ 

Discriminant, $$D=2^2-4\times\ 8\times\ 4<0$$

This has no real solutions for a , and then there will be no real solutions for x as well. 

So x has 4 real solutions, the maximum among them is $$\sqrt{7}$$ and the minimum is $$-\sqrt{7}$$

So their difference is $$\sqrt{\ 7}-\left(-\sqrt{7}\right)=2\sqrt{\ 7}$$ which is option D.