#3**+10 **

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 :))

Melody
Mar 15, 2015

#1**+10 **

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. :)$$$

Melody
Mar 15, 2015

#2**+10 **

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......."

CPhill
Mar 15, 2015

#3**+10 **

Best Answer

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 :))

Melody
Mar 15, 2015