Question

You will get 45 points!!
What is the value of
$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } } }$

Answer 1

Hey there !!


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


▶➡ 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 ✅✅.

____________________________________




$\huge \boxed{ \mathbb{THANKS}}$




$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

Here is your solution

Let


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

$x = \sqrt{6 + x} \\ now \: squaring \: both \: side$
$x {}^{2} = ( \sqrt{6 + x)} {}^{2} \\ x {}^{2} = 6 + x \\ x {}^{2} - x - 6 = 0 \\ x {}^{2} - 3x + 2x - 6 = 0 \\ x(x - 3) + 2(x - 3) = 0 \\ (x + 2)(x - 3)$
=>X+2=0
=>x= -2

=>x-3=0
=>x=3

We will take x = 3 (positive) because root never negative case.

Hence

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

Hope it helps you