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If $x+1= x^{2}$ and $x> 0$, then $2x^{4}$ is

1. $6+4\sqrt{5}$
2. $3+5\sqrt{5}$
3. $5+3\sqrt{5}$
4. $7+3\sqrt{5}$

Given that, $x + 1 = x^{2} \quad \longrightarrow (1)$

$\Rightarrow x^{2}-x-1 = 0 \quad \longrightarrow (2)$

Consider the quadratic equation  $ax^{2} + bx + c = 0$

And the solution of the above equation is  $x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

From the equation $(2),$ we get $a = 1, b = -1,c = 1$

$x = \dfrac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(-1)}}{2\times1}$

$\Rightarrow x = \dfrac{1\pm \sqrt{1+4}}{2}$

$\Rightarrow x = \dfrac{1\pm \sqrt{5}}{2}$

On neglecting the negative value (because $x>0)$, we have

$\Rightarrow \boxed{x = \dfrac{1 + \sqrt{5}}{2}}$

On squaring the equation $(1),$ we get

$({x^{2}})^{2} = (x+1)^{2}$

$\Rightarrow x^{4} = x^{2}+1+2x \quad [\because (a+b)^{2} = a^{2}+2ab+b^{2}]$

$\Rightarrow x^{4} = x + 1 + 1 +2x \quad [\because \text{From equation}\; (1)]$

$\Rightarrow x^{4} = 3x+2$

$\Rightarrow 2x^{4} = 2(3x+2)$

$\Rightarrow 2x^{4} = 6x+4$

$\Rightarrow 2x^{4} = 6\left[\dfrac{1 + \sqrt{5}}{2}\right]+4 \quad [\because \text{Use the value of}\; x]$

$\Rightarrow 2x^{4} = 3(1+\sqrt{5}) + 4$

$\Rightarrow 2x^{4} = 3+3\sqrt{5} + 4$

$\therefore \boxed{2x^{4} = 7+3\sqrt{5}}$

Correct Answer $: \text{D}$
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