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Help rearranging this equation

  • Thread starter skaboy607
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  • #1
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Ive been trying to rearrange this equation for ages and I cant get the answer that is given in the book. I have 0=-4x^2/2+5/2(x-2)^2+16.7. This can be rearranged to give x^2-20x+53.4=0. I cant get anywhere near that, also how would you sove for x from here?

Thanks for your help.
 

Answers and Replies

  • #2
21
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is the first term -(4/2)(x^2) or (-4)(x^(2/2))? In both cases you can easily simplify it
 
  • #3
142
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Ive been trying to rearrange this equation for ages and I cant get the answer that is given in the book. I have 0=-4x^2/2+5/2(x-2)^2+16.7. This can be rearranged to give x^2-20x+53.4=0. I cant get anywhere near that, also how would you sove for x from here?

Thanks for your help.

The way you wrote it is quite confusing, did you mean (all of other stuff written)+16.7 or is that part of the 2(x-2)^2 term... keep things like this in mind though while trying to solve...

say you have .5/1 which equals .5... you can rewrite that as (1/2)/1 .... you can re-arrange this by multiplying by (1/1) to cancel the term on the bottom so you get:

(1/2)*(1/1)= .5 .... im not sure if that helped, but the main point is you can re-arrange divisions by multiplication in a sense.

as far as solving x^2-20x+53.4=0, look up the quadratic equation.
 

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