Question

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

• A. $6+4\sqrt{5}$
• B. $3+5\sqrt{5}$
• C. $5+3\sqrt{5}$
• D. $7+3\sqrt{5}$

Answer 1

The correct option is D

$7+3\sqrt{5}$

$x+1=x^2\Rightarrow x^2-x-1=0\Rightarrow x=\frac{(1+\sqrt{5})}{2}$ given $x>0$
$2x^4\Rightarrow 2x^2(x+1)\Rightarrow 2x^3+2x^2\Rightarrow 2x^2+2x+2x^2\Rightarrow 4x^2+2x\Rightarrow 4(x+1)+2x\Rightarrow 6x+4$
$=3+3\sqrt{5}+4=3\sqrt{5}+7$