Question

find the solution anyone fast ​

Answer 1

Answer :

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

Squaring both sides -

$x {}^{2} = 12 + x \\ \\ x {}^{2} - x - 12 = 0 \\ \\ x {}^{2} - 4x + 3x - 12 = 0 \\ \\ x(x - 4) + 3(x - 4) = 0 \\ \\ (x - 4)(x + 3) = 0 \\ \\ \bf x = 4 \: \: or \: \: x = - 3$

Taking positive value of x,

x = 4

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