If a 0, then sqrt {a + sqrt {a + sqrt {a + | vedclass

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(B) Let $x = \sqrt {a + \sqrt {a + \sqrt {a + .....\infty } } } $Since the expression is infinite,we can write:$x = \sqrt {a + x}$Squaring both sides:

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$\frac{1}{2}[1 + \sqrt {4a + 1} ]$

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(B) Let $x = \sqrt {a + \sqrt {a + \sqrt {a + .....\infty } } } $Since the expression is infinite,we can write:$x = \sqrt {a + x}$Squaring both sides:$x^2 = a + x$Rearranging the terms:$x^2 - x - a = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-a)}}{2(1)}$$x = \frac{1 \pm \sqrt{1 + 4a}}{2}$Since $a > 0$,the value of $x$ must be positive. Therefore,we discard the negative root:$x = \frac{1 + \sqrt{4a + 1}}{2}$

If $a > 0,$ then $\sqrt {a + \sqrt {a + \sqrt {a + ....\infty } } } $ is

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