anonymous
  • anonymous
solve the equation in the set of complex numbers. give only exact answers, not approximations... x^(-2)-x^(-1)=(3/4)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
how would you go about with the negative exponents?
anonymous
  • anonymous
x^(-n)=1/(x^n)
anonymous
  • anonymous
then how would you get rid of fractions?

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anonymous
  • anonymous
\[(2\pm 2\sqrt{2}i)/3\]
anonymous
  • anonymous
how'd you get that..?
anonymous
  • anonymous
solving the equation... :D
anonymous
  • anonymous
nice haha but im still confused on what to do after the x's are on the bottom part of the fractions
anonymous
  • anonymous
make the bottom part x^2 u will get like this: (1-x)/x^2=3/4 after cross multiplication, u can get a quadratic expression, solve that u will get the answer :)
anonymous
  • anonymous
ok so i got to the (1-x)/x^2=3/4 what do i do next? O.o
anonymous
  • anonymous
cross multiplication..
anonymous
  • anonymous
\[x^{-2}-x^{-1}=(3/4)\]\[\Leftrightarrow\frac{1}{x^2}-\frac{1}{x}=\frac{3}{4}\]
anonymous
  • anonymous
agree with dalvoron. Crossed multiply it and get x-x^2=3/4. then transpose 3/4 to the other side:)
anonymous
  • anonymous
am i supposed to get \[1\pm \sqrt{2}i/2\]?
anonymous
  • anonymous
No need for complex numbers here.\[\frac{1}{x^2}-\frac{1}{x}=\frac{3}{4}\]\[\Leftrightarrow \frac{4}{x^2}-\frac{4}{x}=3\]\[\Leftrightarrow 4-4x=3x^2\]
anonymous
  • anonymous
thanks! i just use the quadratic formula and then get both answers :D
anonymous
  • anonymous
Quite right, quite right. Regular factorising is possible also. What did you get as your solutions for x?
anonymous
  • anonymous
2/3,-2
anonymous
  • anonymous
Correct, good job!
anonymous
  • anonymous
THANKS :D
anonymous
  • anonymous
i have a few more problem that i cant get -.-

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