Yes that is what I was doing Chris, I just didn't halve them because I wanted to do it in my head. :))
Here is another way Sweaty, although it is probably over your head.
You can use calculus and get a lot of accuracy very quickly using
Newton's method of approximating roots.
I derive this formula every time I use it for a have a really bad memory but most people just memorise the formula.
$$\boxed{x_2=x_1-\frac{f(x_1)}{f'(x_1)}}$$
$$\\let\;\;x=\sqrt{6}\\
x^2=6\\
x^2-6=0\\
$ the idea is to let $ f(x)=x^2-6 $ and solve for $y=0\\\\
f(x)=x^2-6\\
f'(x)=2x\qquad $ that is calculus $\\
$now we know that $\;\; 2 $So my first estimate will be $x_1=\textcolor[rgb]{1,0,0}{\mathbf{2.5}}\\\\
x_2=2.5-\frac{2.5^2-6}{2*2.5}\\\\
x_2=\textcolor[rgb]{1,0,0}{\mathbf{2.45}}$$
$$\\\mbox{second use of the formula}\\\\
x_3=2.45-\frac{2.45^2-6}{2*2.45}\\\\
x_3=\textcolor[rgb]{1,0,0}{\mathbf{2.449489796}}\\\\
$third use of the formula$\\\\
x_4=2.449489796-\frac{2.449489796^2-6}{2*2.449489796}\\\\
x_4=\textcolor[rgb]{1,0,0}{\mathbf{2.449510072}}\\\\
$The value is already correct to 4 decimal places }
\sqrt{6}=2.4495\\\\\\
$With each iteration the accuracy will increase.$$$
You will learn and understand this when you do calculus - something to look forward too :))
Well there would be a number of ways Sweatie. The easiest one is to use a calculator.
This is another way:
You know it is between 2 and 3
$$2<\sqrt{6}<3$$
$$\\2.5^2=6.25 \qquad $that is too big$\\\\
2 < \sqrt{6}<2.5\\\\
2.3^2=5.29\qquad $too small$\\\\
2.3 < \sqrt{6}<2.5\\\\
2.4^2=5.76 \qquad $too small$\\\\
2.4 < \sqrt{6}<2.5\\\\
etc$$
you can continue to do this untill you get the accuracy that you need.
Note: I did not use a calculator at all for what I have done so far there.
This is an incite into how I do it. :)
$$\\23^2=(20+3)^2=400+2*3*20+9=400+120+9=529\\
so\\
2.3^2=5.29\\
$I can do that in my head - with practice you could too. :)$$$
Here is a method that is similar to Melody's...it converges pretty fast....even if our intial guess is not too good.
This method is known as "Guess - Divide - Check "
Step 1 : Let's say that we guess the square root of 6 as 2
Step 2 : Divide 6 by 2 = 3
Step 3 : Now add 2 +3 and divide by 2 = 2.5
Step 4: Now divide 6 by 2.5 = 2.4
Step 5: Now add 2.5 and 2.4 and divide by 2 = 2.45
Step 6 : Divide 6 by 2.45 = 2.448
Step 7 : Add 2.448 and 2.45 and divide by 2 = 2.449
Step 8: Divide 6 by 2.449 = 2.449......note this is pretty close to the actual square root of 6 !!!
This isn't hard to do.....once you get the "rhythm" down.....!!!!......with a better initial guess.....the procedure may have even converged much more quickly!!!
P.S. - The method might be better termed as "Guess - Divide - Average - Divide - Average - Divide......."
Yes that is what I was doing Chris, I just didn't halve them because I wanted to do it in my head. :))
Here is another way Sweaty, although it is probably over your head.
You can use calculus and get a lot of accuracy very quickly using
Newton's method of approximating roots.
I derive this formula every time I use it for a have a really bad memory but most people just memorise the formula.
$$\boxed{x_2=x_1-\frac{f(x_1)}{f'(x_1)}}$$
$$\\let\;\;x=\sqrt{6}\\
x^2=6\\
x^2-6=0\\
$ the idea is to let $ f(x)=x^2-6 $ and solve for $y=0\\\\
f(x)=x^2-6\\
f'(x)=2x\qquad $ that is calculus $\\
$now we know that $\;\; 2 $So my first estimate will be $x_1=\textcolor[rgb]{1,0,0}{\mathbf{2.5}}\\\\
x_2=2.5-\frac{2.5^2-6}{2*2.5}\\\\
x_2=\textcolor[rgb]{1,0,0}{\mathbf{2.45}}$$
$$\\\mbox{second use of the formula}\\\\
x_3=2.45-\frac{2.45^2-6}{2*2.45}\\\\
x_3=\textcolor[rgb]{1,0,0}{\mathbf{2.449489796}}\\\\
$third use of the formula$\\\\
x_4=2.449489796-\frac{2.449489796^2-6}{2*2.449489796}\\\\
x_4=\textcolor[rgb]{1,0,0}{\mathbf{2.449510072}}\\\\
$The value is already correct to 4 decimal places }
\sqrt{6}=2.4495\\\\\\
$With each iteration the accuracy will increase.$$$
You will learn and understand this when you do calculus - something to look forward too :))