CAT 2017 Slot 1 QA Question & Solution

AlgebraMedium

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

$6+4\sqrt{5}$

$3+3\sqrt{5}$

$5+3\sqrt{5}$

$7+3\sqrt{5}$

We know that $x^2 - x - 1=0$
Therefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
Therefore, $2x^4 = 6x+4$

We know that $x>0$ therefore, we can calculate the value of $x$ to be $\frac{1+\sqrt{5}}{2}$
Hence, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$