sqrt{12+sqrt{12+sqrt{12+cdots}}} = ? | vedclass

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(B) Let the given expression be $x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}$.Squaring both sides,we get $x^2 = 12 + \sqrt{12+\sqrt{12+\cdots}}$.Since th

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let the given expression be $x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}$.Squaring both sides,we get $x^2 = 12 + \sqrt{12+\sqrt{12+\cdots}}$.Since the expression is infinite,the term under the square root is again $x$,so $x^2 = 12 + x$.Rearranging the terms,we get the quadratic equation $x^2 - x - 12 = 0$.Factoring the quadratic equation,we get $(x - 4)(x + 3) = 0$.This gives two possible values for $x$: $x = 4$ or $x = -3$.Since the square root of a positive number must be positive,we neglect $x = -3$.Therefore,$x = 4$.

$\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}} = ?$

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