The solution of the equation x = sqrt{2 + sqr | vedclass

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

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$2$

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(C) Given $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$Squaring both sides,we get $x^2 = 2 + \sqrt{2 + \sqrt{2 + \dots}}$Since the expression inside the square root is $x$,we have $x^2 = 2 + x$Rearranging the terms,we get $x^2 - x - 2 = 0$Factoring the quadratic equation: $(x - 2)(x + 1) = 0$This gives $x = 2$ or $x = -1$Since the square root function must yield a non-negative value,$x 
eq -1$Therefore,$x = 2$.

The solution of the equation $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$ is:

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