Is it possible to find the square root of $2,$ using the method stated here.
I think some other method is needed, but difficult to find explanation, as why it must fail.
Say, find the square root of $2,$ then the square is of dimension: $1+x.$
Next, need add two rectangles of area: $x,$ and one square of area $x^2,$ to area $1,$ in order to get area of $\sqrt{2}^2= 2.$
That is: $1+x^2+2x = 2\implies x^2+2x -1=0.$
So, $x=\frac{-2\pm \sqrt{4+4}}{2} \implies x = -1\pm \sqrt{2}$.
Obviously, $0\lt x \lt 1,$ so, $x = -1+\sqrt{2}$.
But, the question returns to finding $\sqrt 2$, again.
How to reason that this method cannot help in finding square root of an irrational number?
And, why the problem becomes circular, as comes back to the same problem?
To elaborate, if had taken, the task of finding square root of $3$, instead.
Then, again the square is of area: $1$, and the rest area is $2$.
Next, need add two rectangles of area: $x$, and one square of area $x^2$, to area $1,$ in order to get area of $\sqrt{3}^2= 3$.
The equation is: $1+x^2+2x = 3\implies x^2+2x -2=0$.
So, $x=\frac{-2\pm \sqrt{4+8}}{2} \implies x = -1+ \sqrt{3}$.
So, the task is back to finding square root of $3$.
The video shows how to get the digits of a square root by the "division" method. This method generates the digits of the square root one digit at a time.
For the square root of $2$ the method in the video says that $1+x(2+x)\leq 2$. (Note that $x$ must be a digit to the right of the decimal point, that is, it is a multiple of $0.1$.) Your mistake is that you write $=$ instead of $\leq$ and there is no digit that will solve this.
Your next step should be to find the largest multiple of $0.1$ that solves $1+x(2+x)\leq 2$.
Keep in mind that if an integer (such as $2$ or $3$) is not a perfect square, the square root of that number cannot be written exactly with a finite number of decimal digits. But the "division" method will let you find as many digits as you want so that you can make your answer as close to the exact answer as you want.
The video is delightful, by the way.