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Question 22 : If x + 1 = x^{2} and x > 0, then 2x^{4} is :

- 6 + 4√5
- 3 + 5√5
- 5 + 3√5
- 7 + 3√5

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Given that x + 1 = x^{2} and x > 0 then we have to find the value of 2x^{4}.

This 2x^{4} can be found by either finding x first so let us take,

⟹ x + 1 = x^{2}

⟹ x^{2} - x – 1 = 0

⟹ x^{2} - x + \\frac{1}{4}) = 1 + \\frac{1}{4})

⟹ (x - \\frac{1}{2}))^{2} = \\frac{5}{4})

⟹ x - \\frac{1}{2}) = ± \\frac{√5}{2})

⟹ x = \\frac{1}{2}) ± \\frac{√5}{2})

⟹ x + 1 = x^{2}

⟹ (x + 1)^{2} = (x^{2})^{2}

⟹ x^{2} + 2x + 1 = x^{4}

⟹ x + 1 + 2x + 1 = x^{4}

⟹ 3x + 2 = x^{4}

⟹ 6x + 4 = 2x^{4}

We have already found the value of x = \\frac{1}{2}) ± \\frac{√5}{2}), so we can substitute that in place of x. We will get,

⟹ 6(\\frac{1 + √5}{2})) + 4 = 2x^{4}

⟹ 3 + 3√5 + 4 = 2x^{4}

⟹ 7 + 3√5 = 2x^{4}

Hence if x + 1 = x^{2} and x > 0, then 2x^{4} is equal to 7 + 3√5.

Key thing here is to somehow find x, and then write x^{4} in terms of x and then find the value of 2x^{4}

The question is **"If x + 1 = x ^{2} and x > 0, then 2x^{4} is :" **

Choice D is the correct answer

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