Interval bisection to find a root of f(x) I'm attempting to understand Interval bisection. I'm given a simple question in my textbook, and I can do the process easily, I just don't know when to stop. The question is "Use Interval bisection to find the positive root of $x^2 - 7 = 0$, correct to one decimal place" (basically find the square root of 7 to 1 dp)
This is the solution I'm given:

How is it known that it is 2.6?
The last line shows that the root is between 2.640625(from (a+b)/2) and 2.65625(from b).
2.640625 rounds to 2.6 but
2.65625 rounds to 2.7
Surely I would have to keep going until both the upper and lower limit of the interval round to 2.6? 
If it's just simple truncation why didn't the solution stop on the second last line?
(this is just a simple question, so it is as if you can't just do root 7 on a calculator)

I would share your concern that this is not complete.  As David Mitra says, one further iteration is needed. 
In fact since at the third line you know $$2.625^2-7 \lt 0 \lt 2.75^2-7,$$ and so you know the rounded answer is $2.6$ or $2.7$ and could save some effort by checking $$2.65^2-7=+0.0225,$$ so $2.6$ is the rounded solution. 
