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jvaldez45 Group Title

If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    what is the probability that the roots of the equation what?

    • one year ago
  2. satellite73 Group Title
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    what it the probability that the roots of the equation are real?

    • one year ago
  3. jvaldez45 Group Title
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    are both real?

    • one year ago
  4. satellite73 Group Title
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    oh ok we can solve that using the quadratic formula

    • one year ago
  5. jvaldez45 Group Title
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    ok

    • one year ago
  6. jvaldez45 Group Title
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    what would be the b value of the equation. is it a(x+1)

    • one year ago
  7. satellite73 Group Title
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    \[x^2+ax+a+80=0 \] \[x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}\]

    • one year ago
  8. satellite73 Group Title
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    no if you use the quadratic formula, \(a=1,b=a, c=a+80\) different \(a\) of course

    • one year ago
  9. satellite73 Group Title
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    if these are to be real numbers, that means you must have \[a^2-4a-300\geq 0\]

    • one year ago
  10. jvaldez45 Group Title
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    how did you get that?

    • one year ago
  11. satellite73 Group Title
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    by some miracle this factors as \[(a-20)(a+16)\geq 0\] so we can actually solve

    • one year ago
  12. satellite73 Group Title
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    how did i get \(a^2-4a-320\)?

    • one year ago
  13. jvaldez45 Group Title
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    yes

    • one year ago
  14. satellite73 Group Title
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    or how did i get the whole equation?

    • one year ago
  15. phi Group Title
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    how did you get that? the stuff under the square root has to be 0 or positive, else you get imaginary numbers when you take the root.

    • one year ago
  16. jvaldez45 Group Title
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    thanks

    • one year ago
  17. satellite73 Group Title
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    ok quadratic formula tells you the solution to \(ax^2+bx+c=0\) is \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) in your case \(a=1,b=a,c=a+80\)

    • one year ago
  18. satellite73 Group Title
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    when you compute you get \[x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}\] in your case so for these to be real, the discriminant \(a^2-4a-320\) must be greater than or equal to zero, otherwise you have a negative number under the radical

    • one year ago
  19. satellite73 Group Title
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    oh, what @phi said

    • one year ago
  20. jvaldez45 Group Title
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    okay so after i find my values what do I do?

    • one year ago
  21. satellite73 Group Title
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    your last job is to solve for \(a\) \[a^2-4a-320\geq 0\]

    • one year ago
  22. jvaldez45 Group Title
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    i found my values and they are a=20 and a+-16

    • one year ago
  23. satellite73 Group Title
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    this factors as \((a-20)(a+16)\geq 0\)

    • one year ago
  24. jvaldez45 Group Title
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    a=-16

    • one year ago
  25. satellite73 Group Title
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    hold on, you are not solving \((a-20)(a+16)=0\) but rather \[(a-20)(a+16)\geq 0\]

    • one year ago
  26. satellite73 Group Title
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    the zeros are \(-16\) and \(20\) for sure but you want to know the interval over which it is positive since this is a quadratic with leading coeffient positive, it will be positive outside the zeros, in other words if \(a\leq -16\) or \(a\geq 20\)

    • one year ago
  27. satellite73 Group Title
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    your final job is to find what portion of the interval \([-28,28]\) satisfies \(a\leq -16\) or \(a\geq 20\)

    • one year ago
  28. jvaldez45 Group Title
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    okay so i would have to find the interval of the it by integrating it?

    • one year ago
  29. satellite73 Group Title
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    not necessary, just look

    • one year ago
  30. satellite73 Group Title
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    favorable part is \([-28,-16]\cup [20,28]\)

    • one year ago
  31. satellite73 Group Title
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    you can eyeball the total length

    • one year ago
  32. jvaldez45 Group Title
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    12 and 8

    • one year ago
  33. satellite73 Group Title
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    right, for a total of 20

    • one year ago
  34. jvaldez45 Group Title
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    thanks just found the answer. thank you

    • one year ago
  35. satellite73 Group Title
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    yw

    • one year ago
  36. satellite73 Group Title
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    hope it was \(\frac{20}{56}\)

    • one year ago
  37. jvaldez45 Group Title
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    yea

    • one year ago
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