The value of sqrt{42+sqrt{42+sqrt{42+ldots}}} | vedclass

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Question

(A) Let $y = \sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$.Since the expression is infinite,we can write:$y = \sqrt{42+y}$.Squaring both sides,we get:$y^2 = 4

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let $y = \sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$.Since the expression is infinite,we can write:$y = \sqrt{42+y}$.Squaring both sides,we get:$y^2 = 42 + y$.Rearranging the terms to form a quadratic equation:$y^2 - y - 42 = 0$.Factoring the quadratic equation:$(y - 7)(y + 6) = 0$.This gives two possible values for $y$: $y = 7$ or $y = -6$.Since the square root of a positive number must be positive,$y$ cannot be $-6$.Therefore,the value is $y = 7$.

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

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