Question

★100 points ★ Challenge. ✌✌☺☺

Prove that:

${ \boxed{ \sf \therefore \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } = 3.}}$
​

If you can.

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Answer 1

Answer :-

→ 3 .

Step-by-step explanation :-

[ Note :- This question can be solved by two methods ] .

$\huge \pink{ \underline{ \sf 1st \: Method }}$

[ For competitions exams ; short method ]

→ Check the number which is in under root i.e., 6 .

→ Then, do the prime factorisation of the number that you found under root i.e., 6 = 2 × 3 .

→ Thus, the greatest of the prime factors is the answer of this type questions i.e., 3 .

$\orange{ \boxed{ \sf \therefore \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } = 3.}}$

$\huge \pink{ \underline{ \sf 2nd \: Method }}$

[ For board exams ; long method ]

$\begin{lgathered}\begin{lgathered}\sf Let \: x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.... \infty } } } } . \\ \\ \sf \implies x = \sqrt{6 + \bigg( \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } \bigg) } \\ \\ \sf \implies x = \sqrt{6 + x} . \\ \\ \{ \tt squaring \: both \: side \} \\ \\ \sf \implies {x}^{2} = 6 + x. \\ \\ \sf \implies {x}^{2} - x - 6 = 0. \\ \\ \sf \implies {x}^{2} - 3x + 2x - 6 = 0. \\ \\ \sf \implies x(x - 3) + 2(x - 3) = 0. \\ \\ \sf \implies (x + 2)(x - 3) = 0. \\ \\ \sf \implies x + 2 = 0. \: \: \green{or} \: \: x - 3 = 0. \\ \\ \sf \implies x = 3 \: \: \green{or} \: \: x = - 2.\end{lgathered}\end{lgathered}$

[ °•° Since, x is in under root , then x must be positive i.e., x = 3 ]

$\orange{ \boxed{ \sf \therefore \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } = 3.}}$

Hence, it is solved.

Answer 2

→ 3 .

Step-by-step explanation :-

[ Note :- This question can be solved by two methods ] .

[ For competitions exams ; short method ]

→ Check the number which is in under root i.e., 6 .

→ Then, do the prime factorisation of the number that you found under root i.e., 6 = 2 × 3 .

→ Thus, the greatest of the prime factors is the answer of this type questions i.e., 3 .