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Finding the perfect square

  • Thread starter Miike012
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  • #1
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how do you find the perfect square of say

ax2 + b/x2 + c
??
 

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  • #2
Mentallic
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You can't find the perfect square of that problem precisely, unless you're satisfied with [tex]\left(\sqrt{ax^2+\frac{b}{x^2}+c}\right)^2[/tex] which I doubt since it's trivial, but take a look at the expansion of

[tex]\left(x+\frac{1}{x}\right)^2[/tex]
 
  • #3
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how do you find the perfect square of say

ax2 + b/x2 + c
??
You didn't by chance mean (ax2 + b)/(x2 + c), did you? If so, the lack of parentheses around the numerator and denominator completely confused Mentallic about what you're asking.
 
  • #4
Mentallic
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You didn't by chance mean (ax2 + b)/(x2 + c), did you? If so, the lack of parentheses around the numerator and denominator completely confused Mentallic about what you're asking.
That possibility completely skipped my mind :biggrin:
 
  • #5
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Mentallic,
Well, I'm about as puzzled by this problem as you must be. The way I read it, the OP just wants to square the original expression, whatever it is.
 
  • #6
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Nope that is what I ment to say.. I added an example to the paint doc and highlighted the portion in red.

It has to do with finding the surface area of a curve... and basically I was unaware the equation could be turned into a perfect square... so was wondering if there is some pattern I should look for ?
 

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  • #7
Mentallic
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Mentallic,
Well, I'm about as puzzled by this problem as you must be. The way I read it, the OP just wants to square the original expression, whatever it is.
I didn't have any doubts about what the OP is trying to do, just what the expression was meant to be once you raised the point.


Mike, like I was saying it doesn't work in general that ax2 + b/x2 + c can be turned into a perfect square, but in this case c happened to be the right number for the job.

When you get to the expression

[tex]\frac{25}{36}x^8+\frac{1}{2}+\frac{9}{100}x^{-8}[/tex]

You should realize that it could be of the form [tex]\left(ax^4+bx^{-4}\right)^2[/tex] where in this case [tex]a=\sqrt{\frac{25}{36}}=\frac{5}{6}[/tex]
[tex]b=\sqrt{\frac{9}{100}}=\frac{3}{10}[/tex]

And all you'd need to do is check to see if [tex]2\cdot \frac{5}{6}\cdot \frac{3}{10} =\frac{1}{2}[/tex]
 

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