Question

find the value of x =

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\textbf{\underline{Answer}}$ = 3

$\textbf{\underline{Explanation}}$ :-

Since,

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

so, we can take x = $\sqrt{6 + x}$ because x is equal to the whole term !

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

Now, squaring both sides :

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

square root and square will cancel out!

$= > {x}^{2} = 6 + x$

then, we have to take any value of x such that it's square should be equal to 6 + the value of x.

$\therefore$ (3)² = 6 + 3

=> 9 = 9

thus , we get LHS = RHS .