Question

★ Challenge ★. ✌✌☺☺

Prove that:

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

If you can.

No spam.​​

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 ]

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

[ °•° 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

Question :-

Prove that

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

Solution :-

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

Now square both sides

$x {}^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } }$

We know that

$x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6.... \infty } } } }$

Hence,

$x {}^{2} = 6 + x \\ \\ x {}^{2} - x - 6 = 0$

Now Factorise it by splitting middle term

$x {}^{2} - 3x + 2x - 6 = 0 \\ \\ x(x - 3) + 2(x - 3) = 0 \\ \\ (x - 3)(x + 2) = 0$

Now split it into possible cases

$x_1 = (3 )\\ \\ x_2 =( - 2)$

Because in the question we have positive sign

Hence,

The value we take will be in positive = (3)

Proved