Sum of Infinitely Many Square Roots

Let $x$ be the given number. Then:

$\begin{aligned} x & = \sqrt{6+\color{#D61F06} \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}} \\ x & = \sqrt{6+\color{#D61F06}x} & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 & = 6 + x \\ x^2-x-6 & = 0 \\ (x-3)(x+2) & = 0 \\ \implies x & = \boxed 3 & \small \color{#3D99F6} \text{Note that }x > 0 \end{aligned}$

$\sqrt{6+ \sqrt{6+ \sqrt{6+ \sqrt{6+ \sqrt{6+ ...} } } } } = a$

$6+ \sqrt{6+ \sqrt{6+ \sqrt{6+ \sqrt{6+ ...} } } } = a^{2}$

$\sqrt{6+ \sqrt{6+ \sqrt{6+ \sqrt{6+ ...} } } } = a^{2} - 6$

$a = a^{2} - 6$

$a^{2} - a - 6 = 0$

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

$a = -2, 3$

But as $a$ can't be negative, as it is a square root,

$a = 3$

Let x=√6+(√6+(√6...)

Now, x=√(6+x)

=> x^2 = 6 + x

x^2 - x - 6 = 0

Solving the above equation (not shown) will give

x = $\frac{1±5}{2}$

As there is no negative answer in the options

$\frac{1+5}{2}$ =3

Hence the answer is 3

Well let's go step by step 1. Let us write it down as (√ 6+√ 6+√ 6+√ 6+√ 6+√ 6+...........∞) = a 2. now let us consider the fact that the number of square roots of six that we are supposed to add is infinite as in endless. 3. lets convert it this way (a² = 6+√ 6+√ 6+√ 6+√ 6+√ 6+...........∞) because when we multiply a by itself on the LHS, we have to remove a root on the other side. 4. now as in step 1, we know that the sum of infinitely many square roots of six are denoted by 'a'. 5. so in step 3, after '6+' we again have infinitely many square roots of six which are again denoted by 'a'. 6. so now we can reform the question as a² = 6 + a. 7. now it is pure algebra. so let us solve. 8. a² = 6+a = a x a - a = 6 = a = 3 3 x 3 - 3 = 6 9 - 3 = 3 3 = 3 LHS = RHS