Question

Solve this..
$\tt{} \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3+ \sqrt{3} ...} } } } } = x$
find the value of x ?​

Answer 1

3 is the answer

Let x =

$\tt{} \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3+ \sqrt{3} ...} } } } } = x$

On squaring both sides,

$\sf \: x {}^{2} = 6 + \sqrt{6 \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6} } } } }$

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

$\sf \: x {}^{2} −x−6=0$

$\sf \: x = \frac{1 + \sqrt{1 {}^{2} - 4 \times 1( - 6) } }{2(1)}$

$\sf \: x = \frac{1 + \sqrt{25} }{2} = \frac{1 + 5}{2}$

$\sf \: ∴x=3 or x=−2$

Since , a negative number cannot be under root, hence x = 3