#cube roots

https://youtu.be/HxjIdydOpGI

This Is How We Math: Square and Cube Roots

*making middle school math fun* :)

Last week, one of the topics we explored was square and cube roots. (Super fun!)

To give Grace a nice hands-on experience, I came up with this little idea:

Yep, Grace made squares and cubes to investigate this whole square and cube root thing. :D

So, here’s what we did:

For each measurement (2″, 3″, 4″, 5″, and 6″), Grace cut out 7 squares. (Bonus: Practice with…

Submitted by Sav2718:

I saw that someone wrote as a comment on my business card: "I saw something similar for fifth roots- getting the last digit right is a lot easier too!". So I just wanted to add a word about the rightmost digit of a perfect root in general: The pattern for the rightmost digit is the same for every order of the form 4K+1 (the digit itself) and for 4K-1 (the same unique pattern as with cube roots). the same is true for roots of an even-order but an extra step of elimination is needed, two steps for even-order of the form 4K. For example, lets take a look at some (perfect) square roots: 1^2=1 2^2=4 3^2=9 4^2=16 5^2=25 6^2=36 7^2=49 8^2=64 9^2=81 10^2=100 11^2=121 12^2=144 13^2=169 14^2=196 15^2=225 16^2=256 17^2=289 18^2=324 19^2=361 20^2=400 etc. the pattern for the rightmost digit is always (0,)1,4,9,6,5,6,9,4,1(,0) 1 for 1 and 9 (the complement to 10) 4 for 2 and 8 9 for 3 and 7 6 for 4 and 6 5 for 5 a trail of 2N 0s (even number) in the right hand of the number beneath the root sign will be a trail of N 0s in right hand of the answer (ex: sqrt(15210000)=3900). Now in order to calculate a given perfect square root lets look at an example: sqrt(5329) 70^2(=4900)<5329<80^2(=6400) or even more simplified: 7^2(=49)<53<8^2(=64) so the leftmost digit of the answer is 7 5329 ends with a 9 so according to our pattern the rightmost digit is either 3 or 7 but how can we tell? Let's have a look at the trick for squaring a number that ends with 5: (10a+5)^2=100a^2+2*10*5*a+5^2=100a^2+100a+25=100a*(a+1)+25 or even more simplified: [a*(a+1)]&[25] (where "&" is the concatenation operator) examples: 35^2=[3*4]&[25]=[12]&[25]=1225 75^2=[7*8]&[25]=[56]&[25]=5625 435^2=[43*44]&[25]=[1892]&[25]=189225 you may also prefer to calculate it as such: 35^2=[3^2+3]&[25]=[9+3]&[25]=[12]&[25]=1225 75^2=[7^2+7]&[25]=[49+7]&[25]=[56]&[25]=5625 435^2=[43^2+43]&[25]=[1849+43]&[25]=[1892]&[25]=189225 Now back to our square root: we know sqrt(5329) is either 73 or 77 but we can quickly check that 75^2=[7*8]&[25]=[56]&[25]=5625>5329 so if 75 is too big, 77 is even bigger than the answer must be (correctly) 73. Note that because a square of number ending with 5 always ends with 25 one doesn't even need to calculate the whole thing and just have a look at 7*8 (7 times its successor). Lets do another one quickly: sqrt(107584) 32^2(=1024)<1075 (don't need the '84') <33^2(=1089) 32_ 107584 ends with 4 so the rightmost digit is either 2 or 8 32^2+32=1024+32=1056<1075 too small so 322 is even smaller so the answer is (correctly) 328. As I already mentioned even-order roots from the form 4K need another step of elimination.  Higher numbers require more steps and non-perfect roots require algorithms from a different kind and approach but I am willing to write about those in the future. 

Submitted by Sav2718:

New business card sample. The back is an algorithm for calculating cube roots mentally and\or on paper (up to 6-digit numbers, higher numbers require extra steps and lookup tables) with examples.