Question 26

Let $$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$$. Then x equals

Solution

$$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$$

=> $$x = \sqrt{4+\sqrt{4-x}}$$

=> $$x^2 = 4 + \sqrt{4-x}$$

=>$$x^4 + 16 - 8x^2 = 4 - x$$

=> $$x^4 - 8x^2 + x +12 = 0$$

On substituting options, we can see that option C satisfies the equation.

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