Question

Solve this plsssssssssss​

Answer 1

$\sqrt{5 + \color {red} \sqrt{5 + \sqrt{5 + \sqrt{5 + ...} } } } = x$

Observe that,

$\color{red} \sqrt{5 + \sqrt{5 + \sqrt{5 + ...} } } \color{black} = x$

Putting the value in original equation,

$\sqrt{5 + x} = x$

Squaring Both Sides,

$5 + x = {x}^{2} \\ \implies {x}^{2} - x - 5 = 0$

By Using the Quadratic Formula,

$x = \frac{ 1 \pm \sqrt{ {( - 1)}^{2} - 4( 1)( - 5)} }{2(1)}$

$x = \frac{1 \pm \sqrt{21} }{2}$

As, $x$ tends to a positive value,

$x = \frac{1 + \sqrt{21} }{2}$

Answer 2

As, xx tends to a positive value,

x = \frac{1 + \sqrt{21} }{2}x=21+21