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(A) Given $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{1 + x}$.Squaring both sides,we

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$\frac{1 + \sqrt{5}}{2}$

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(A) Given $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{1 + x}$.Squaring both sides,we get $x^2 = 1 + x$.Rearranging the terms,we obtain the quadratic equation $x^2 - x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -1, c = -1$:$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.Since $x$ represents a square root,$x$ must be positive $(x > 0)$.Therefore,we discard the negative root $\frac{1 - \sqrt{5}}{2}$.Thus,$x = \frac{1 + \sqrt{5}}{2}$.

If $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$,then $x =$

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