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anonymous

  • 5 years ago

x^3 -2x^2+13x=0 leave in exact form no decimals approx

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  1. anonymous
    • 5 years ago
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    You could write the same equation as: x(x^2 - 2x + 13) = 0 so one root is x = 0 and the others are the roots of x^2 - 2x + 13 =0 equation which are (1 - 2sqrt3) i and (1 + 2sqrt3 i)

  2. anonymous
    • 5 years ago
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    did you solved this using quadratic formula negative b + or - square root b^2 - 4ac over 2a?

  3. radar
    • 5 years ago
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    I believe he did and he simplified the answer.

  4. anonymous
    • 5 years ago
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    i tried that i did broke it down i got 2+ or - sqrt-48 over 2 times "a" which is 2 but when you divide the top (2+ or - sqrt-48) i got 24 which breaks down to to 2i sqrt 6 how did he get the 3?

  5. radar
    • 5 years ago
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    \[\sqrt{-48}=\sqrt{16\times-3}\]

  6. radar
    • 5 years ago
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    \[\sqrt{16}\times \sqrt{-3}\]\[4\sqrt{-3}\]

  7. radar
    • 5 years ago
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    does that simplification rings a bell?

  8. anonymous
    • 5 years ago
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    yeah i get it now its hard when someone just leaps from one spot to another

  9. radar
    • 5 years ago
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    I haven't double checked but nikola usually gets it right. Good luck

  10. anonymous
    • 5 years ago
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    Sweets, http://en.wikipedia.org/wiki/Quadratic_equation a=1 b=(-2) c=13 b^2 - 4ac = (-2)^2 - 4*1*13 = -48 which is smaller than 0,so roots are complex and can be calculated as : -b/2a + (sqrt [- (b^2 - 4ac)] /2a)*i -b/2a - (sqrt [- (b^2 - 4ac)] /2a)*i

  11. anonymous
    • 5 years ago
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    yay nikola lol

  12. radar
    • 5 years ago
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    And I have to agree with you sometimes simplified looks more complicated lol\[\sqrt{-48}=4\sqrt{-3}\] doesn't look to much more simple Hi

  13. anonymous
    • 5 years ago
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    2+(sqrt[-2^2-4(1)(13)]/2(1) = 2+(sqrt[-48])/2 = so do I divide out the 2 from the bottom like so or no lol sorry for all this... 1+(sqrt[-24] then you break down the 24 to 4i sqrt 6

  14. anonymous
    • 5 years ago
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    so 1+\[4i \sqrt{6} and 1-4i \sqrt{6}\]

  15. anonymous
    • 5 years ago
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    that was suppose to be 2i sqrt6 not the 4 my bad

  16. anonymous
    • 5 years ago
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    Sweets, a=1 b=(-2) c = 13 b^2 - 4ac = (-2)^2 - 4*1*13 = -48 < 0 so; -b/2a + (sqrt [- (b^2 - 4ac)] /2a)*i -b = -(-2) = 2 so -b/2a is 2/2*1 =1 sqrt [- [b^2 - 4ac) ] = sqrt [-(-48)] = sqrt48 which is 4sqrt3 (sqrt [- (b^2 - 4ac)] /2a)*i is 4sqrt3 / 2 *1 = 2sqrt3 I hope this helps.

  17. anonymous
    • 5 years ago
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    i have got 2+24i and 2-24i as the answer

  18. anonymous
    • 5 years ago
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    yes thats correct nikola:)

  19. anonymous
    • 5 years ago
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    Sweets,it s -b/2*a not -b/a :)

  20. radar
    • 5 years ago
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    \[(-(-2)\pm \sqrt{(-2)^{2}-(4)(1)(13)})/2\]\[(2\pm \sqrt{4-52})/2\]\[(2\pm \sqrt{-48})/2\] can you take it from there?

  21. anonymous
    • 5 years ago
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    Oooooo lol i see i see

  22. anonymous
    • 5 years ago
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    you miss one thing it messes it all up lol

  23. anonymous
    • 5 years ago
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    yeah...thats what math is all about:)

  24. anonymous
    • 5 years ago
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    haha gotta love it

  25. anonymous
    • 5 years ago
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    so th eanswer probably is 1+24i & 1-24i if i am not wrong:)

  26. radar
    • 5 years ago
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    Continuing on.\[(2\pm(\sqrt{16\times-3})/2\] \[(2\pm(\sqrt{16}\sqrt{-3})/2\]\[(2\pm(4\sqrt{-3})/2\] now do the division\[1\pm2\sqrt{-3}\]

  27. anonymous
    • 5 years ago
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    how did u get -3 there? under the root???

  28. radar
    • 5 years ago
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    I would leave it there are you could go further and do this. Convert the radical as follows:\[\sqrt{-3}=\sqrt{3 X-1}\]

  29. radar
    • 5 years ago
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    The square root of -1 is the imaginary operator i so the final answer becomes:\[1\pm \sqrt{3}i\]

  30. radar
    • 5 years ago
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    Did you understand I was trying to show in the radical 3 times a -1 ? and i left out the 2 so it should be\[1\pm2\sqrt{3}i\]

  31. anonymous
    • 5 years ago
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    yeah i seen what you did 4 squared times 3 is 48 but you had to divide the 4 by 2 i get it

  32. radar
    • 5 years ago
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    Yeahh!!!Now practice is the key. Notice that they did not want you to really get the value no decimals etc. so that is as far as you need to take it.

  33. anonymous
    • 5 years ago
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    lol thank all of you so much. Im not as slow as i seem this was just a total brain fart

  34. anonymous
    • 5 years ago
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    Im a visual learner so it helped when you showed :)

  35. anonymous
    • 5 years ago
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    you're welcome sweets,good luck with your studies.

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