sqrt{6+sqrt{6+sqrt{6+cdots}}} is equal to | vedclass

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(D) Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$.Squaring both sides,we get:$x^{2} = 6 + \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$Since the expression inside

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$3$

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(D) Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$.Squaring both sides,we get:$x^{2} = 6 + \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$Since the expression inside the square root is the same as $x$,we can substitute $x$:$x^{2} = 6 + x$Rearranging the terms to form a quadratic equation:$x^{2} - x - 6 = 0$Factoring the quadratic equation:$x^{2} - 3x + 2x - 6 = 0$$x(x - 3) + 2(x - 3) = 0$$(x + 2)(x - 3) = 0$This gives $x = 3$ or $x = -2$.Since the square root of a positive number must be positive,$x$ cannot be $-2$.Therefore,$x = 3$.

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is equal to

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