Question

Solve the above question​

Answer 1

Answer:

(b) 3

Step-by-step explanation:

Let $\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} = x$

The nested radical inside from the second term onwards can also be replaced by $x$ as it is an infinite series

$\sqrt{6+x} = x$

Square both sides

$6+x = x^2$

$(x-3)(x+2) = 0$

$x = 3~ \text{or} ~-2$

But $x$ cannot be negative as it is equal to a square root

$\therefore x =3$

Answer 2

Need to FinD :-

• The value of $\sf\sqrt{ 6 + \sqrt{ 6 + \sqrt{6 + ... . .\infty }}}$

$\red{\frak{Given}}\Bigg\{ \sf \sqrt{ 6 + \sqrt{ 6 + \sqrt{6 + ... . .\infty }}}$

Let us take that the value of given nested radical is x . Therefore ,

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

As here we can see that term $\sqrt{ 6 + \sqrt{ 6 + \sqrt{6 + ... . .\infty }}}$ repeats itself inside the squareroot . Therefore we can write it as ,

$\sf\longrightarrow x = \sqrt{ 6 + x }$

On squaring both sides and simplyfing , we have ,

$\sf\longrightarrow x^2 = 6 + x \\\\\\\sf\longrightarrow x^2 - x - 6 = 0 \\\\\\\sf\longrightarrow x^2 -3x +2x - 6 = 0 \\\\\\\sf\longrightarrow x( x - 3 ) +2( x -3) = 0 \\\\\\\sf\longrightarrow ( x -3)( x +2) = 0 </p><p>\\\\\\\sf\longrightarrow \underline{\underline{\red{ x = 3 \ , \ (-2) }}}$

On ignoring the negative Value , we have,

$\sf\longrightarrow \underline{\underline{\red{\textsf{\textbf{ x = 3 \ \ }}}}}$