If x = sqrt{2 + sqrt{2 + sqrt{2 + dots}}},the | vedclass

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(C) Given the expression $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$.Squaring both sides,we get $x^2 = 2 + \sqrt{2 + \sqrt{2 + \dots}}$.Since the inn

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$x = 2$

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(C) Given the expression $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$.Squaring both sides,we get $x^2 = 2 + \sqrt{2 + \sqrt{2 + \dots}}$.Since the inner part is the same as $x$,we can write $x^2 = 2 + x$.Rearranging the terms,we get the quadratic equation $x^2 - x - 2 = 0$.Factoring the equation: $(x - 2)(x + 1) = 0$.This gives $x = 2$ or $x = -1$.Since $x$ is the square root of a positive value,$x$ must be positive.Therefore,$x = 2$.

If $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$,then find the value of $x$.

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