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Root equation help

  • Thread starter thomas49th
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  • #51
arildno
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That's impossible.
you have most likely miscopied the original problem.
 
  • #52
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Ok, must of done. Now interestly the next question on the paper is

Hence or otherwise sove

[tex] \frac{1}{x-2} + \frac{2}{x+4} = \frac{1}{3}[/tex]

so I will EXPAND the fractions (or cross multiply)?
this will give me

9x = x² - 2x + 4x - 8
so this is a quadratic. - 9x

x² -7x + 8 = 0
(x + 1)(x-8) = 0
so x = -1 or x= 8

that correct?
 
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  • #53
arildno
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Which is a totally different issue altogether!
What you have there is an EQUATION, what you said before was that that equality was an IDENTITY (which is NOT correct).
 
  • #54
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aha i see
to solve it then

[tex] \frac{1}{x-2} + \frac{2}{x+4} = \frac{1}{3}[/tex]

so I will EXPAND the fractions (or cross multiply)?
this will give me

9x = x² - 2x + 4x - 8
so this is a quadratic. - 9x

x² -7x + 8 = 0
(x + 1)(x-8) = 0
so x = -1 or x= 8

that correct?
 
  • #55
arildno
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Seems so, yes.
 
  • #56
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O, right first time...

Now here's a hard one I dont get

Find the value of

m when [tex]\sqrt{128} = 2^{m}[/tex]

I no straight away from binary that 2^7 is 128 does that help?
 
  • #57
arildno
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Indeed it helps!
Remember how roots can be written as exponents..
 
  • #58
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[tex]\frac{7}{2}[/tex]

but how would i solve it normally?
 
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  • #59
arildno
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Indeed, that is what m equals, as soon as you get the LateX right..:smile:
 
  • #60
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but say if i didn't know about binary how would i got about solving it
somthing to do with surds isn't it?
 
  • #61
HallsofIvy
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In fact, perhaps your original post contained a typo. You went from
A = πr² + r²root(k²-1) to r³ = (2A/π+root(k²-1)
Did you really mean the r³, rather than r²?
 
  • #62
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:redface: i believe your right
 

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