determine between which twonconsecutive whole numbers each square root lies and estimate the irrational number to the nearest tenth

Square root of 12

3^2=9   4^2=16  so it is about half way between between  3 and 4.  If I estimated at this point I'd say 3.5

3.5^2=12.25  that is too big but very close

3.45^2=11.9025  to small

So to one decimal place it is 3.5

You can do the others yourself.

Best way of doing this sort of thing is to use Newton's method.

The formula  for finding the square root of the number A is,

$\displaystyle x_{n+1}=\frac{1}{2}\left(x_{n} + \frac{A}{x_{n}}\right)$

where $\displaystyle x_{n}$

is your current approximation

and $\displaystyle x_{n+1}$

is your new approximation.

So, if your first approximation for the square root of 12 is 3.5, your next approximation will be

(3.5 + 12/3.5)/2 = 3.46 (2dp).

Next would be

(3.46 + 12/3.46)/2 = 3.4641 (4dp).

And the next

(3.4641 + 12/3.4641)/2 = 3.4641016 (7dp).

The number of correct decimal places roughly doubles at each step.

(So if 3.5 is correct to 1dp, then 3.46 is probably correct to 2 dp, and so on.)

Easiest is to put each current approximation into memory and then recall it when needed, (i.e you wouldn't enter 3.46 at the second step, you'd put the result of the calculation (3.5 + 12/3.5)/2 into memory and then use that).