Distance from cartesian coordinates and im going wring somewhere.

  • Thread starter EMFsmith
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  • #1
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Why is

[tex]\sqrt{9+36}[/tex]
= [tex]3\sqrt{5}[/tex]

and not 6.708 ?

I wasnt interested in maths at school but now i'm trying to self teach, so pardon my ignorance.

Edit: Ok i feel foolish now, no need to correct me as ive just worked out i WAS correct. I'm still unsure as to why it would be shown like this though.
 
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Answers and Replies

  • #2
danago
Gold Member
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Edit: Ok i feel foolish now, no need to correct me as ive just worked out i WAS correct. I'm still unsure as to why it would be shown like this though.
6.708 is only an approximation, whereas [itex]3\sqrt{5}[/itex] is exact. If you put [itex]3\sqrt{5}[/itex] into a calculator you will see that it does not stop after three decimal places. For some calculations where maximum accuracy is required, it may be inappropriate to round, in which case the square root notation should be used.
 
  • #3
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6.708 is only an approximation, whereas [itex]3\sqrt{5}[/itex] is exact. If you put [itex]3\sqrt{5}[/itex] into a calculator you will see that it does not stop after three decimal places. For some calculations where maximum accuracy is required, it may be inappropriate to round, in which case the square root notation should be used.
Thats great thanks, I'm struggling to remember how to work out the square root notation instead of a decimal?
 
  • #4
danago
Gold Member
1,122
4
Thats great thanks, I'm struggling to remember how to work out the square root notation instead of a decimal?
[tex]\sqrt{9+36}=\sqrt{45}=\sqrt{9\times 5}=\sqrt{9}\times \sqrt{5}=3\sqrt{5}[/tex]

Hopefully that helps
 
  • #5
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[tex]\sqrt{9+36}=\sqrt{45}=\sqrt{9\times 5}=\sqrt{9}\times \sqrt{5}=3\sqrt{5}[/tex]

Hopefully that helps
Yeah thats great thanks!
 
  • #6
Mentallic
Homework Helper
3,798
94
[tex]\sqrt{9+36}[/tex]
= [tex]3\sqrt{5}[/tex]
Let me guess, the hypotenuse of a right angled triangle with sides 3 and 6? :wink:
 
  • #7
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Let me guess, the hypotenuse of a right angled triangle with sides 3 and 6? :wink:
Could well be the lenghts of the other 2 sides, I was going from the question

P(-2,3) Q(1,-3)

Find the lenght between the 2 points.

First attempt at this so the advice has been really useful, sometimes text books just dont do it for me.
 
  • #8
Mentallic
Homework Helper
3,798
94
Could well be the lenghts of the other 2 sides, I was going from the question

P(-2,3) Q(1,-3)

Find the lenght between the 2 points.

First attempt at this so the advice has been really useful, sometimes text books just dont do it for me.
The formula you used was derived from pythagoras' theorem. If you draw a right-angled triangle with the hypotenuse being PQ then you'll see it's a right-angled triangle with side lengths 3 and 6.
 

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