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August 1st, 2007, 02:28 PM   #1
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Square roots question

What is the square root of 19 to the nearest whole number?

What is the square root of 171 to the nearest whole number?

Solve F to the second power = 169

is square root of 36 rational, irrational, interger rational, or other?

estimate b to the 2nd power = 525 to the nearest integer.

square root of 17 to nearest tenth.

square root of negative 102 round to the nearest tenth..

thanks.. if these could be answered by today that would help out and everything.


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August 1st, 2007, 04:30 PM   #2
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let us assume that we can't use any calculator related tools, and let us solve it both algebraically and in calculus terms.

Q. What is the square root of 19 to the nearest whole number?
A. let us assume that newton's method will help us solve this. let
f(x)=x^2-19=0, where we want to find for x, approximation. take the derivative of f(x).. which gives us f'(x)=2x. let us assume that x_0=4, and x_1=5. let's use the linear approximation, using general equation
f(x)=l(x)=f(x_k)+f'(x_k)(x-x_k), where k is the kth term of x.
let's use x_0=4 first.
f(x)=l(x)=f(4)+f'(4)(x-4)=-3+8(x-4)=8x-35=0
8x=35
x=35/8=4.375 is an example of an close approximation.
let's use the x_1=5 for our final approximation.
f(x)=l(x)=f(5)+f'(5)(x-5)=6+10(x-5)=10x-44=0
10x=44
x=44/10=22/5=4.4 is an another example of an close approximation.

since the 4.375 is an approximation for x_0=4, and 4.4 is an approximation for x_1=5, and x=sqrt(19) must be between those two approximations, we can say that nearest whole number for sqrt(19) is 4.
do this the same way for your second problem, third problem, fifth, sixth and seventh problem.

since sqrt(36)=6, we can say that it's "positive, integer, rational, even, natural, composite number".
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August 11th, 2007, 09:40 AM   #3
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Quote:
Originally Posted by johnny
Q. What is the square root of 19 to the nearest whole number?
A. let us assume that Newton's method will help us solve this.
Why make that assumption? Newton's method may fail for problems that are only slightly more difficult.

What is your answer for the third problem?

Also, do you really do things exactly the same way for the seventh problem?
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August 11th, 2007, 12:44 PM   #4
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Quote:
Why make that assumption? Newton's method may fail for problems that are only slightly more difficult.

What is your answer for the third problem?

Also, do you really do things exactly the same way for the seventh problem?
Since the math problem here is generally according to basic arithmetic, is it better idea to use some basic simplifying and calculating results, instead of using Newton's Method?
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August 11th, 2007, 02:27 PM   #5
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Re: Help.. Please

I did some mental math to give pointers to the answers.

Quote:
Originally Posted by Time
What is the square root of 19 to the nearest whole number?
For positive reals, the square root increases monotonically, so it's clear that the answer must be 4 or 5 (since 16 < 19 < 25). Thus all you have to see is whether 4.5 * 4.5 is more or less than 19.

Quote:
Originally Posted by Time
What is the square root of 171 to the nearest whole number?
Same as above, but with 13 and 14. Actually your intuition should give you this one right away without calculation, if you don't need a proof.

Quote:
Originally Posted by Time
Solve F to the second power = 169
There's a negative and a positive answer. This is easy because the answer is a whole number.

Quote:
Originally Posted by Time
is square root of 36 rational, irrational, interger rational, or other?
This is easy: square roots of integers other than perfect squares are always transcendental, so all you have to see is if 36 is a perfect square or not.

Quote:
Originally Posted by Time
estimate b to the 2nd power = 525 to the nearest integer.
Again, two answers. Off the top of my head 23 is close, but check that this is the closest.

Quote:
Originally Posted by Time
square root of 17 to nearest tenth.
4.0 < x < 4.4, since 16 < 17 < 1.21 * 16. Just subdivide the intervals a few more times, checking each time to see what side 17 is on.

Quote:
Originally Posted by Time
square root of negative 102 round to the nearest tenth..
10.1 squared is easy to calculate, since you can just treat it as a binomial. (10+0.1)^2 = 100 + 2*10*0.1 + 0.01 = 102.01. I imagine that's the answer.
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August 12th, 2007, 03:25 PM   #6
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Re: Help.. Please

Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by Time
square root of negative 102 round to the nearest tenth..
10.1 squared is easy to calculate, since you can just treat it as a binomial. (10+0.1)^2 = 100 + 2*10*0.1 + 0.01 = 102.01. I imagine that's the answer.
Be careful with that one. The question asked about the square root of a negative number. I assume all these questions are over the set of reals, because there is no real sensible way to define nth root over the complex numbers that will make it not multivalued. Over the reals, the square root of a negative number is undefined.

Over the set of complex numbers, the answer would be ±10.01i, where i:=√-1.
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August 13th, 2007, 07:54 AM   #7
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Re: Help.. Please

Quote:
Originally Posted by roadnottaken
Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by Time
square root of negative 102 round to the nearest tenth..
10.1 squared is easy to calculate, since you can just treat it as a binomial. (10+0.1)^2 = 100 + 2*10*0.1 + 0.01 = 102.01. I imagine that's the answer.
Be careful with that one. The question asked about the square root of a negative number. I assume all these questions are over the set of reals, because there is no real sensible way to define nth root over the complex numbers that will make it not multivalued. Over the reals, the square root of a negative number is undefined.

Over the set of complex numbers, the answer would be ±10.01i, where i:=√-1.
No, just 10.01i; it asked for the square root of -102, not the solutions for the equation x * x = -102.
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August 13th, 2007, 08:40 AM   #8
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I suppose you can stil define square root as a function over complex numbers. The difficulty arises in odd roots, where the most natural definition over complex numbers disagrees with the most natural definition over real numbers when you examine the negative reals.

For example, when dealing with roots of complex numbers, the most natural definition for the cube root of -1 (if you insist on making cube root a function) is cbrt(e^(pi*i))=e^(pi*i/3)=1/2+sqrt(3)*i/2, where as, over the set of reals, the only candidate for the cube root of -1 is -1.

That is why I think it makes more sense to, once you start dealing with complex numbers, let nth root be multivalued (specifically, it has n values, as long as n is an integer).
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August 13th, 2007, 08:47 AM   #9
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Quote:
Originally Posted by roadnottaken
For example, when dealing with roots of complex numbers, the most natural definition for the cube root of -1 (if you insist on making cube root a function) is cbrt(e^(pi*i))=e^(pi*i/3)=1/2+sqrt(3)*i/2, where as, over the set of reals, the only candidate for the cube root of -1 is -1.
I've always seen it defined as a function, never as a multivalued function. Of course conventions vary, but the ones I've seen choose the root as the one in the appropriate arc of the plane starting at the real axis and moving as far as it needs to -- so theta is in [0, 1/n) where 0 is the real line and the interval moves counterclockwise.

But yes, if it's multivalued you should have two answers, and if the range is defined as the reals you should have zero. Maybe the teacher should tell students how many answers to give...
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