If x = sqrt {1 + sqrt {1 + sqrt {1 + dots inf | vedclass

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Question

(A) Given the expression $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can substitute the inner part with $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the expression $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots \infty}}}$.Since the expression is infinite,we can substitute the inner part with $x$,so $x = \sqrt{1 + x}$.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 $a = 1, b = -1, c = -1$,we get:$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.Since $x$ represents a square root of a positive sum,$x$ must be greater than $0$.Therefore,we discard the negative root $\frac{1 - \sqrt{5}}{2}$ because it is less than $0$.Thus,$x = \frac{1 + \sqrt{5}}{2}$.

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

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