Question

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$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6...... \infty } } } }$
find the value

Answer 1

$\bold{\huge{\purple{\underline{\mathcal{\star\:Hello\:Dude\:\star}}}}}$


$let \: \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6........} } } } be \: x \\ so \\ x = \sqrt{6 + x} \\ squaring \: both \: sides \: \\ {x}^{2} = 6 + x \\ {x}^{2} - x - 6 = 0 \\ (x - 3)(x + 2) = 0 \: \: \: \: \: \: (on \: factorising) \\ x = 3 \: or \: x = - 2 \\ but \: we \: know \: that \: \sqrt{} \: \: can \: never \\ give \: a \: negative \: value \: so \: \\ x = 3 \: is \: the \: correct \: answer$

Answer 2

$\huge \boxed{ \mathbb{ANSWER:}}$


▶➡ Find the value:-)


$\bf \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } } = ?$


This question can be solved by two methods .


$\huge \boxed{1st \: Method.}$
[For competition exam (short process ) ].


↪➡ See the number which is in the under root. i.e., 6.

↪➡ Now, Find the factors of the number that is in the under root whose difference is 1.

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

↪➡ Therefore, The greatest factors is the answer of this question. i.e., 3.



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




$\huge \boxed{ 2nd \: Method.}$
[ For board exam ( long Process ) ].



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


$\bf => x = \sqrt{6 + x} .$

[ Squaring both side ].



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


=> x² = 6 + x.

=> x² - x - 6 = 0.

=> x² - 3x + 2x - 6 = 0.

=> x ( x - 3 ) + 2 ( x - 3 ) = 0.

=> ( x + 2 ) ( x - 3 ) = 0.

=> x + 2 = 0. | x - 3 = 0.

=> x = -2. | x = 3.


↪ [ Root is never in the negative case ].

▶ So, x = 3.


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


✔✔ Hence, it is solved ✅✅.

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