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If $a>0$, then $\sqrt {a+\sqrt {a+\sqrt{a+............}}}$=?

$\begin{array}{1 1}(A)\large\frac{\sqrt{4a-1}}{2} \\ (B) \large\frac{1+\sqrt{4a+1}}{2} \\ (C) \large\frac{1-\sqrt{4a-1}}{2} \\(D) \large\frac{1\pm\sqrt{4a+1}}{2} \end{array}$

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1 Answer

Let $\sqrt{a+\sqrt{a+\sqrt{a+.................}}}=x$
Squaring both the sides we get
$a+\sqrt{a+\sqrt{a+..................}}=x^2$
$\Rightarrow\:x^2-a=\sqrt{a+\sqrt{a+\sqrt{a+.......}}}=x$
$\Rightarrow\:x^2-a=x$
$\Rightarrow\:x^2-x-a=0$
$\Rightarrow\:x=\large\frac{1\pm {\sqrt{1+4a}}}{2}$
answered Jul 27, 2013 by rvidyagovindarajan_1
 

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