Question

plzz solve the question ​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let

$y = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + .......} } } }$

$= > y^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ....} } }$

$= > {y}^{2} = 6 + y$

$= > {y}^{2} - y - 6 = 0$

$= > {y}^{2} - 3y + 2y - 6 = 0$

$= > y(y - 3) + 2(y - 3) = 0$

$= > (y - 3)(y + 2) = 0$

Since, square root is a positive quantity, so, y ≠ -ve

$= > y = 3$