roots in roots in roots in roots forever

$a=\sqrt{12+\sqrt{12+\sqrt{12+...}}}$

$a^{2}=12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}$

$a^{2}=12+a$

$a^{2}-a-12=0$

$(a-4)(a+3)=0$

The negative solution will be disregarded since the expression was originally positive.

$\boxed{a=4}$

$\displaystyle \sqrt{12+\sqrt{12+\sqrt{12+...... }}}$

or $\displaystyle \sqrt{12+x}= x$

or, $\displaystyle 12+x=x^{2}$

or, $\displaystyle x^{2}-x-12=0$

or, $\displaystyle (x+3)(x-4)=0$

or, $\displaystyle x=4$

We do not take the negative root of the equation because the terms of the expression are all positive and hence their sum can never be a negative value.

Let $S$ be $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12…}}}}$ , so: $S^2=(\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12…}}}})^2$ $S^2=12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12…}}}}$

$\Longrightarrow$ $S^2=12+S$

$\Longrightarrow$ $S^2-S-12=0$

$\Longrightarrow$ $(S-4)(S+3)=0$

$\Longrightarrow$ $S=4$ or $S=-3$

We take the positive value 'cause the terms are positive, then their sum is positive

$\sqrt{12 + \sqrt{12+ \sqrt{12+ \sqrt{12.....}}}}$ = $x$

$(\sqrt{12 + \sqrt{12+ \sqrt{12+ \sqrt{12....}}}}$ ) $^{2}$ = $x^{2}$

$(12 + \sqrt{12 + \sqrt{12+ \sqrt{12}}}$ = $x^{2}$

$(12 + x)$ = $x^{2}$

0 = $x^{2}$ - $x$ - $12$

0 = $(x - 4) (x+3)$

Therefore $x = 4$

$\sqrt {12 + \sqrt {12 + \sqrt {12 +...}}}=\sqrt {12 +x}=x$

$x^2-x-12=0,\,x_{1,2}=\frac {1\pm\sqrt{1+48}} 2,\,x_1=4,\,x_2=-3,\,\Rightarrow\, x= 4,\,$

because $x>0$

I took a more intuitive approach, which is probably not recommended, but works here. I'm assigning a chain of variables at random, to make the discussion more straightforward.

Starting under the lowest nested radical, we can set

$z = 12$

take the square root of both sides

$\sqrt{z} = \sqrt{12}$

the square root of 12 is between the square roots of 9 and 16,

Thus, $3 < \sqrt{z}\ < 4$ ;

add 12 to $\sqrt{z}$ , and call it $y$

$y = 12 + \sqrt{z}$

$y$ is greater than 15, but less than 16, so . . .

$3 < \sqrt{y} < 4$

Which is the same inequality as \sqrt{z}\

It's a quick assessment to realize that even if you take it all the way to $a = 12 + \sqrt{b}...$ , you'll never reach or exceed 4, so the limit of the function is 4.

suppose, $x=\sqrt{12+\sqrt{12+\sqrt{12+.....}}}$

then, $x=\sqrt{12+x}$

or, $x^2=12+x$

doing middle term we get, $x=4,-3$

but square root of positive number can not be negative.

so, $x=4$

$\sqrt {12+ \sqrt {12+ \sqrt {12...}}}=S$ $12+ \sqrt {12+ \sqrt {12+ \sqrt {12...}}}=S+12$ $\sqrt {12+ \sqrt {12+ \sqrt {12+ \sqrt {12...}}}}=\sqrt {S+12}$ So:

$\sqrt {S+12}=S$

$S+12=S^2$

$S^2-S-12=0$

$(S+3)(S-4)=0$

$S=4$

Like Anik Mandal said, we do not take the negative root of the equation because the terms of the expression are all positive and hence their sum can never be a negative value.

$\begin{aligned} \sqrt { 12+\sqrt { 12+\sqrt { 12+\sqrt { 12+\dots } } } } & \overset { let }{ = } & x \\ \sqrt { 12+x } & = & x \\ 2+x & = & { x }^{ 2 } \\ { x }^{ 2 }-x-12 & = & 0 \\ (x-4)(x+3) & = & 0 \\ x & = & 4\quad (x>0) \\ \sqrt { 12+\sqrt { 12+\sqrt { 12+\sqrt { 12+\dots } } } } & = & \boxed { 4 } \end{aligned}$