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# How to find the approximate value of sprt(6)

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How to find the approximate value of sprt(6)

Mar 15, 2015

#3
+95360
+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.

$$\boxed{x_2=x_1-\frac{f(x_1)}{f'(x_1)}}$$

$$\\let\;\;x=\sqrt{6}\\ x^2=6\\ x^2-6=0\\  the idea is to let  f(x)=x^2-6  and solve for y=0\\\\ f(x)=x^2-6\\ f'(x)=2x\qquad  that is calculus \\ now we know that \;\; 2 So my first estimate will be x_1={\mathbf{2.5}}\\\\ x_2=2.5-\frac{2.5^2-6}{2*2.5}\\\\ x_2={\mathbf{2.45}}$$

$$\\\mbox{second use of the formula}\\\\ x_3=2.45-\frac{2.45^2-6}{2*2.45}\\\\ x_3={\mathbf{2.449489796}}\\\\ third use of the formula\\\\ x_4=2.449489796-\frac{2.449489796^2-6}{2*2.449489796}\\\\ x_4={\mathbf{2.449510072}}\\\\ The value is already correct to 4 decimal places } \sqrt{6}=2.4495\\\\\\ With each iteration the accuracy will increase.$$$You will learn and understand this when you do calculus - something to look forward too :)) Mar 15, 2015 ### 4+0 Answers #1 +95360 +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. :) $$\\23^2=(20+3)^2=400+2*3*20+9=400+120+9=529\\ so\\ 2.3^2=5.29\\ I can do that in my head - with practice you could too. :)$$$

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Mar 15, 2015
#2
+94558
+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......."

Mar 15, 2015
#3
+95360
+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.

$$\boxed{x_2=x_1-\frac{f(x_1)}{f'(x_1)}}$$

$$\\let\;\;x=\sqrt{6}\\ x^2=6\\ x^2-6=0\\  the idea is to let  f(x)=x^2-6  and solve for y=0\\\\ f(x)=x^2-6\\ f'(x)=2x\qquad  that is calculus \\ now we know that \;\; 2 So my first estimate will be x_1={\mathbf{2.5}}\\\\ x_2=2.5-\frac{2.5^2-6}{2*2.5}\\\\ x_2={\mathbf{2.45}}$$

$$\\\mbox{second use of the formula}\\\\ x_3=2.45-\frac{2.45^2-6}{2*2.45}\\\\ x_3={\mathbf{2.449489796}}\\\\ third use of the formula\\\\ x_4=2.449489796-\frac{2.449489796^2-6}{2*2.449489796}\\\\ x_4={\mathbf{2.449510072}}\\\\ The value is already correct to 4 decimal places } \sqrt{6}=2.4495\\\\\\ With each iteration the accuracy will increase.$$\$

You will learn and understand this when you do calculus - something to look forward too :))

Melody Mar 15, 2015
#4
+95360
0

Thanks Chris :))

Mar 15, 2015