If a 0,then the value of sqrt{a + sqrt{a + | vedclass

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(B) Let $x = \sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$.Squaring both sides,we get $x^2 = a + \sqrt{a + \sqrt{a + \dots \infty}}$.Since the expres

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

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(B) Let $x = \sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$.Squaring both sides,we get $x^2 = a + \sqrt{a + \sqrt{a + \dots \infty}}$.Since the expression inside the square root is the same as $x$,we can write $x^2 = a + x$.Rearranging the terms,we get the quadratic equation $x^2 - x - a = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-1, c=-a$,we get $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-a)}}{2(1)}$.This simplifies to $x = \frac{1 \pm \sqrt{1 + 4a}}{2}$.Since $x$ must be positive (as it is a square root),we take the positive root: $x = \frac{1 + \sqrt{4a + 1}}{2}$.

If $a > 0$,then the value of $\sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$ is:

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