Question

14- Find a decimal approximation to the nearest tenth:
i) √82

Is there ANY easy way other than:

Find the perfect squares that is before and after the number 82 and then add the both numbers and blah..blah..

Answer 1

since it is to the "nearest tenth" you won't have many numbers that you need to try.
start with 9 since 9*9=81 and increase by one tenth until you get an answer that is closest to 82.

Answer 2

Can you show me?

Answer 3

another possible method might be to use the fact that:\[(a+b)^2=a^2+2ab+b^2\]so set a=9 and you need to find b such that:\[(9+b)^2=82\]

Answer 4

therefore:\[(9+b)^2=81+2*9*b+b^2=81+18b+b^2\]therefore:\[81+18b+b^2=82\]therefore:\[b^2+18b=1\]then try b=0.1, 0.2, ....

Answer 5

Oh cool =) Can you do this method if they asked you " Find a decimal approximation to the nearest hundredth" ? @asnaseer

Answer 6

yes, then you would need to try b=0.01, 0.02, etc. However, this might take too long, so a better strategy in this case might be to use a technique where you keep halving the interval, e.g.:
try b=0.50 ---> if this is too low, then try b=0.75
---> otherwise try b=0.25

that will converge quicker.

Answer 7

have you heard of the Newton-Raphson method?

Answer 8

if you have, then that will converge to the answer faster.

Answer 9

Actually no I havent

Answer 10

no worries - just use one (or more) of the techniques mentioned above to see which one you prefer.

Answer 11

Thank you so much =D

Answer 12

yw :)

Answer 13

Okay, I got a bit stuck using your method.. My method is SOOOOO complicated. Anyway, right you said you try b=0.1,0.2.. like this:

b^2+18b=1
0.1^2=18(0.1)=1 ? and if it didnt work you try 0.2 and if didnt work you try 0.3 and so on? O.o

Answer 14

@asnaseer :)

Answer 15

you won't find an exact answer - you are supposed to find the nearest answer to one tenth

Answer 16

Yeah and I think you have to round it at the end right?

Answer 17

ok, let me try and explain. we said we can re-arrange the question to this form:\[b^2+18b=1\]

Answer 18

but my question is you substitute b with 0.1 like that? until you find an answer to the nearest tenth?

Answer 19

so what we need to do is to find what value of b will give an answer that is "closest" to 1. we can re-arrange further to get:\[b^2+18b-1=0\]so now we need to find a value for b that gives the closest answer to zero.

so, lets try b=0.0, we get \(b^2+18b-1=-1\)
so, lets try b=0.1, we get \(b^2+18b-1=0.81\)
so, lets try b=0.2, we get \(b^2+18b-1=2.64\)

so the "closest" answer is when b=0.1

therefore, the answer is 9.1 to nearest tenth.

Answer 20

does that make sense?

Answer 21

It makes sense till so the "closest" answer is when b=0.1
and then how did it become 9.1?

Answer 22

because we started this from:\[(9+b)^2=82\]now we have found 'b', answer is '9+b'

Answer 23

remember we said \(9^2=81\) and we are trying to find a number that, when squared equals 82. so that number must be 9 plus a little bit. we called that "little bit" 'b'

Answer 24

This is even more complicated lol but yeah i understood how =) Thank youu!!

Answer 25

Afwan :)