Question

pls answer...............

Answer 1

$\huge \bf{HEY \: FRIENDS!!}$

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$\huge \bf \boxed{Here \: is \: your \: answer↓}$

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▶⏩ Fin the value:-)

$\huge{ \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6...... \infty } } } } } .}$

$\huge \boxed{1st \: method}$

↪➡ First, see the number that is in the under root . i.e., 6.

↪➡ After that, find the factor of 6 whose difference is 1.

↪➡ Hence, 2 and 3 is the factors of 6 and
=> 3 - 2 = 1.

↪➡ Therefore, the greatest factor is the answer. i.e., 3.

$\bf = > \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6...... \infty } } } } } = 3.$

$\huge \boxed{2nd \: method}$

$\bf \: let \: x \: = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } .$

$\bf \: x = \sqrt{(6 + x)} .$

$\huge \boxed{Squaring \: both \: side}$

$\bf = > {x}^{2} = { \sqrt{(6 + x)} }^{2} .$

$\bf = > {x}^{2} = 6 + x.$

$\bf = > {x}^{2} - x - 6 = 0.$

$\bf = > {x}^{2} - 3x + 2x - 6 = 0.$

$\bf = > x(x - 3) + 2(x - 3) =0.$

$\bf = > (x - 3)(x + 2) = 0.$

$\huge \bf{ = > x = 3 \: \: and \: \: x = - 2.}$

↪➡ Since, square root is always in positive case. So, answer is 3.

$\bf = > \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } } } = 3.$

✅✅ Hence, it is finded✔✔.

$\huge \boxed{THANKS}$

$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$