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CAT 2017 [slot 1] Question with solution 22

Question 22:
If $x+1=x^{2}$ and $x>0$, then $2x^{4}$ is
  1. $6+4\sqrt{5}$
  2. $3+3\sqrt{5}$
  3. $5+3\sqrt{5}$
  4. $7+3\sqrt{5}$
Option: 4
Explanation:

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, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$


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