Question

if x= root 12 + root 12 + root 12+ till infinity then find the value of x

Answer 1

Answer:

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

$\sqrt{12+x}=x$

$(\sqrt{12+x})^2=(x)^2$

$12+x=x^2$

$x^2-x-12=0$

$x^2+3x-4x-12=0$

$x(x+3)-4(x+3)=0$

$(x-4)(x+3)=0$

$x= 4, -3$

$@rohangupta0424$

Answer 2

Answer:

Not defined

Step-by-step explanation:

Like when we divide a number by zero, this calculation is not possible as the sum goes on and there is not a limit to the numbers (though the universal limit is 1 Gogool [ 1 and a hundred zeroes])

So the value of x can be written as ♾️ or 'not defined'.