SOLUTION: "Between which 2 consecutive inetgers is the square root of 471 located?" I dont know how to find the square root for such a large number. I also dont know how to do the first p

Algebra ->  Radicals -> SOLUTION: "Between which 2 consecutive inetgers is the square root of 471 located?" I dont know how to find the square root for such a large number. I also dont know how to do the first p      Log On


   

Question 187664: "Between which 2 consecutive inetgers is the square root of 471 located?"
I dont know how to find the square root for such a large number. I also dont know how to do the first part. Please help.
Thank you
Found 2 solutions by Earlsdon, jim_thompson5910:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Start by writing down a few squares of numbers:
20%5E2+=+400
21%5E2+=+441
22%5E2+=+484
You can see by inspection that the given number of 471 lies between 441 and 484, so this indicates that the sqrt%28471%29 lies between sqrt%28441%29+=+21 and sqrt%28484%29+=+22.
The answer is 21 and 22

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since we know that 2%5E2=4, this means that 20%5E2=400 (just add 2 zeros)


So this means that 21%5E2=21%2A21=441 and 22%5E2=22%2A22=484 (use either a calculator or multiply out by hand)


Since 441%3C471%3C484, this means that sqrt%28441%29%3Csqrt%28471%29%3Csqrt%28484%29 (apply the square root to each side) which then becomes 21%3Csqrt%28471%29%3C22


So the square root of 471 is in between the integers 21 and 22.


Note: which means that this number is greater than 21 but less than 22.