Question

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

Answer 1

Required Answer:-

Given to evaluate:

• $\sf \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } }$

Solution:

Let us assume that,

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

Squaring both sides, we get,

$\sf \implies {x}^{2} = 6 +\sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } }$

As x = √(6+√(6+...∞), So,

$\sf \implies {x}^{2} = 6 + x$

$\sf \implies {x}^{2} - x - 6= 0$

$\sf \implies {x}^{2} - 3x + 2x- 6= 0$

$\sf \implies x(x - 3) + 2(x - 3)= 0$

$\sf \implies (x + 2)(x - 3)= 0$

Therefore, either x + 2 = 0 or x - 3 = 0

$\sf \implies x = - 2, 3$

But x cannot be negative. So,

$\sf \implies x = 3$

★ Hence, the sum of the nested infinite radical is 3.

Answer:

• Result is: 3.

Answer 2

Answer:

Required Answer:-

Given to evaluate:

\sf \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } }

Solution:

Let us assume that,

\sf \implies x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } } }

Squaring both sides, we get,

As x = √(6+√(6+...∞), So,

\sf \implies {x}^{2} = 6 +\sqrt{6 + \sqrt{6 + \sqrt{6..... \infty } } }