sqrt{6+sqrt{6+sqrt{6+cdots}}} = ? | vedclass

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Question

(B) Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$Since the expression is infinite,we can write:$x = \sqrt{6+x}$Squaring both sides:$x^2 = 6+x$Rearrangin

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$Since the expression is infinite,we can write:$x = \sqrt{6+x}$Squaring both sides:$x^2 = 6+x$Rearranging into a quadratic equation:$x^2 - x - 6 = 0$Factoring the quadratic equation:$(x-3)(x+2) = 0$This gives $x = 3$ or $x = -2$.Since the square root of a positive number must be positive,we ignore $x = -2$.Therefore,$x = 3$.

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} = ?$

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