Question

6
$6 = \sqrt{x + \sqrt{x + \sqrt{x + ..... + \infty \: \: } } }$
find the value of x​

Answer 1

Given:-

• $\tt{6 = \sqrt{x + \sqrt{x + \sqrt{x + ..... + \infty \: \: } } } }$

To Find:-

• The value of x .

Answer:

Given that , $\tt{6 = \sqrt{x + \sqrt{x + \sqrt{x + ..... + \infty \: \: } } } }$.

Now here we can see that in squareroot $\tt{\sqrt{x +\sqrt{x + \sqrt{x + ....}}}}$ is repeating .

So , instead of this we can just substitute 6 , which is given equal to it .

So , the new equⁿ looks something like this :-

$\tt{\implies 6 = \sqrt{x + 6}. }$

Now square both sides in order to remove the squareroot.

$\tt{\implies 6^2 = (\sqrt{x+6})^2.}$

$\tt{\implies 36 = x + 6 .}$

$\tt{\implies x = 36-6.}$

$\underline{\boxed{\red{\bf{\leadsto x = 30.}}}}$