Question

Please help me out to this question.....

Answer 1

answer is 7 dude...........

Answer 2

Let

$\sqrt{ 4 + 3\sqrt{ 4 + 3\sqrt{ 4 + 3\sqrt{4 + ...} } } } = x \\ \sqrt{4 + 3x} = x \\ squaring \: both \: sides \: we \: get : \\ 4 + 3x = {x}^{2} \\ {x}^{2} - 3x - 4 = 0$

Now just solve this quadratic equation:
x² - 3x - 4 = 0
x² - 4x + x - 4 = 0
x(x - 4) + 1(x - 4) = 0
(x-4)(x+1) = 0
So, x = 4, -1
Now, since x = √(...), which can't be negative, so x = 4

Thankyou!!!