Question

please friends, explain me
step by step...


I hope you can​

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\bf{2.}$

$\sf{ \implies\sqrt{2 + \sqrt{2 + \sqrt{2 + \infty } } } = x}$

squaring both side .

$\sf{ \implies ( { \sqrt{2 + \sqrt{2 + \sqrt{2 + \infty } } } })^{2} } \: = x ^{2}$

$\sf{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \infty } } } } = {x}^{2}$

→ 2 + x = x²

→ 0 = x² - x - 2

→ 0 = x² - 2x + x - 2

→ 0 = x(x - 2) + 1 (x - 2 )

→ 0 = (x - 2 ) (x + 1)

→ x - 2 = 0

→ x = 2 .

and

→ x + 1 = 0

→ x = -1

(-1 will be rejected because it is in negative )

Hence

$\huge \sf \pink{2. \: answer}$

Answer 2

▶ Question :-

→ Simplify :-

$\sf \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ... \infty } } } } .$

▶Answer :-

→ 2 .

▶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., 2 .

→ Then, do the prime factorisation of the number that you found under root i.e., 2 = 2 × 1 .

→ Thus, the greatest of the prime factors is the answer of this type questions i.e., 2 .

$\orange{ \boxed{ \sf \therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2... \infty } } } } = 2.}}$

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

[ For board exams ; long method ]

$\begin{lgathered}\sf Let \: x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2.... \infty } } } } . \\ \\ \sf \implies x = \sqrt{2 + \bigg( \sqrt{2 + \sqrt{2 + \sqrt{2... \infty } } } \bigg) } \\ \\ \sf \implies x = \sqrt{2 + x} . \\ \\ \{ \tt squaring \: both \: side \} \\ \\ \sf \implies {x}^{2} = 2 + x. \\ \\ \sf \implies {x}^{2} - x - 2 = 0. \\ \\ \sf \implies {x}^{2} - 2x + x - 2 = 0. \\ \\ \sf \implies x(x - 2) + 1(x - 2) = 0. \\ \\ \sf \implies (x - 2 )(x + 1) = 0. \\ \\ \sf \implies x - 2 = 0. \: \: \green{or} \: \: x + 1 = 0. \\ \\ \sf \implies x = 2 \: \: \green{or} \: \: x = - 1.\end{lgathered}$

[ °•° Since, x is in under root , then x must be positive i.e., x = 2 ]

$\orange{ \boxed{ \sf \therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2... \infty } } } } = 2.}}$

Hence, it is solved.