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anonymous

  • one year ago

In the equation x + kx+ 54 one root is equal to twice the other root. The value(s) of k is (are): a) 5.2 15.6 -5.2 15.6 22.0

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  1. caozeyuan
    • one year ago
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    Well, x+kx+54 is polynomial, not equation, so where is the equation?

  2. anonymous
    • one year ago
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    x + kx+ 54=0

  3. caozeyuan
    • one year ago
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    Cool, now let's do this

  4. caozeyuan
    • one year ago
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    the first thing I would do is solve for x as a function of k

  5. caozeyuan
    • one year ago
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    Wait a sec! this is first order, meaning it's got only one root. are you missing a square?

  6. anonymous
    • one year ago
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    oh sry, x^2

  7. caozeyuan
    • one year ago
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    So we have 2 eqa now.\[x ^{2}+kx+54=0\]

  8. caozeyuan
    • one year ago
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    and \[4x ^{2}+2kx+54=0\]

  9. caozeyuan
    • one year ago
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    with me so far?

  10. anonymous
    • one year ago
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    why are there 2 equations

  11. caozeyuan
    • one year ago
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    Becuase it says one rt is twice the other, so if one rt is x, the other has to be 2x, plug 2x in into the original eqa, we can get a new eqa that is equally valid

  12. anonymous
    • one year ago
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    oh ok

  13. caozeyuan
    • one year ago
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    now divide the new eqa by 4 on both side, give me the result

  14. anonymous
    • one year ago
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    x^2+kx/2 +27/2

  15. caozeyuan
    • one year ago
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    fantastic, subtract this eqa above from the original one, what do you get?

  16. anonymous
    • one year ago
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    why?

  17. caozeyuan
    • one year ago
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    Because you can get rid of the annoying x squared now

  18. anonymous
    • one year ago
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    @caozeyuan sorry, i had to do something. so now i set them equal to eachother and solve for x?

  19. caozeyuan
    • one year ago
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    NONONO!

  20. caozeyuan
    • one year ago
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    you subtract!

  21. radar
    • one year ago
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    The roots are:\[-3\sqrt{3}, -6\sqrt{3}\] . This means:\[k=9\sqrt{3}\] Now to verify that: \[(x + 6 \sqrt{3}) (x + 3\sqrt{3}) = x ^{2}+ 9\sqrt{3}+54\] \[(x +6\sqrt{3})=0\]\[x=-6\sqrt{3}\]\[x = -3\sqrt{3}\]

  22. caozeyuan
    • one year ago
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    @radar, show steps don't just give answers

  23. anonymous
    • one year ago
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    wait, shouldn't the second equation be 2x^2+2kx+54?

  24. caozeyuan
    • one year ago
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    well, no. if x is 2x, then x^2 is (2x)^2 which is 4x^2

  25. anonymous
    • one year ago
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    oh right, so i get kx/2 +24 after subtracting

  26. anonymous
    • one year ago
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    =0

  27. anonymous
    • one year ago
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    then what??

  28. caozeyuan
    • one year ago
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    I got kx+81=0, WTF is going on?!

  29. anonymous
    • one year ago
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    oh i forgot it's 27/2

  30. anonymous
    • one year ago
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    then what

  31. caozeyuan
    • one year ago
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    then it sucks, cuz our answer is not in the choices

  32. anonymous
    • one year ago
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    ..................

  33. anonymous
    • one year ago
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    @phi @Greg_D @mathmate

  34. caozeyuan
    • one year ago
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    My brain is rusty, I haven't had any math class for half an year

  35. anonymous
    • one year ago
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    the answer found by @radar is \(9\sqrt{3}\approx 15.59\) which is close to \(15.6\), the second choice

  36. caozeyuan
    • one year ago
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    But what's wrong w/ my way?! IDC anyway cuz I finished highschool algebra a long time ago

  37. anonymous
    • one year ago
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    Howw?

  38. anonymous
    • one year ago
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    @caozeyuan please give me a minute to read the entire post...

  39. anonymous
    • one year ago
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    i dont see any problems starting with \(P(x)=0\) and \(P(2x)=0\), but when you substract you may be missing a solution, because the \(x^2\) term dissapears

  40. caozeyuan
    • one year ago
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    I see, so I guess I have to solve the eqa with quadratic formula then find k?

  41. anonymous
    • one year ago
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    let me try something in paper, if it works i will post it... just 2 minutes please

  42. anonymous
    • one year ago
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    Maybe we have to use sum and product rules? C/a and -b/a

  43. anonymous
    • one year ago
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    @caozeyuan exactly that. if we solve directly we get: \[x_{\pm} = \frac{-k\pm \sqrt{k^2-216}}{2}\] so we can try either: \(x_+ = 2x_-\) or \(x_- = 2x_+\)

  44. caozeyuan
    • one year ago
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    yep, that's my gut feeling actually, but I though subtraction is much easier

  45. anonymous
    • one year ago
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    @caozeyuan when dealing with quadratic eqs. (or higher order) you have to be careful when you substract and/or divide, you may loose solutions!! now, if we go with: \(x_+=2x_-\) we get: \[\frac{-k+\sqrt{k^2-216}}{2}=-k-\sqrt{k^2-216}\] so \[k=3\sqrt{k^2-216}\] or \[k^2=9(k^2-216)\] which solves to: \(k=-9\sqrt{3}\approx -15.58\) and \(k=9\sqrt{3}\approx 15.58\) so i guess the answer is the second choice \(k=15.6\)

  46. caozeyuan
    • one year ago
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    cool, I got that, but I don't care LOL

  47. anonymous
    • one year ago
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    lol

  48. radar
    • one year ago
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    Sorry about that. I was using the factor method to solve your quadratic equation. The two factors would be of the form (x +a)(x+b) The two factors when multiplied would have to equal your equation\[x ^{2}+kx+54=0\]The product of the numerical terms would have to equal 54. And as stated by the problem b would equal 2b ("one root is equal to twice the other root" and a + 2b would have to equal k.

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