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jvaldez45

  • 2 years ago

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

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  1. satellite73
    • 2 years ago
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    what is the probability that the roots of the equation what?

  2. satellite73
    • 2 years ago
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    what it the probability that the roots of the equation are real?

  3. jvaldez45
    • 2 years ago
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    are both real?

  4. satellite73
    • 2 years ago
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    oh ok we can solve that using the quadratic formula

  5. jvaldez45
    • 2 years ago
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    ok

  6. jvaldez45
    • 2 years ago
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    what would be the b value of the equation. is it a(x+1)

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

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

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

  10. jvaldez45
    • 2 years ago
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    how did you get that?

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

  12. satellite73
    • 2 years ago
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    how did i get \(a^2-4a-320\)?

  13. jvaldez45
    • 2 years ago
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    yes

  14. satellite73
    • 2 years ago
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    or how did i get the whole equation?

  15. phi
    • 2 years ago
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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.

  16. jvaldez45
    • 2 years ago
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    thanks

  17. satellite73
    • 2 years ago
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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 \(a=1,b=a,c=a+80\)

  18. satellite73
    • 2 years ago
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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

  19. satellite73
    • 2 years ago
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    oh, what @phi said

  20. jvaldez45
    • 2 years ago
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    okay so after i find my values what do I do?

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

  22. jvaldez45
    • 2 years ago
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    i found my values and they are a=20 and a+-16

  23. satellite73
    • 2 years ago
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    this factors as \((a-20)(a+16)\geq 0\)

  24. jvaldez45
    • 2 years ago
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    a=-16

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

  26. satellite73
    • 2 years ago
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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 \(a\geq 20\)

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

  28. jvaldez45
    • 2 years ago
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    okay so i would have to find the interval of the it by integrating it?

  29. satellite73
    • 2 years ago
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    not necessary, just look

  30. satellite73
    • 2 years ago
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    favorable part is \([-28,-16]\cup [20,28]\)

  31. satellite73
    • 2 years ago
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    you can eyeball the total length

  32. jvaldez45
    • 2 years ago
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    12 and 8

  33. satellite73
    • 2 years ago
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    right, for a total of 20

  34. jvaldez45
    • 2 years ago
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    thanks just found the answer. thank you

  35. satellite73
    • 2 years ago
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    yw

  36. satellite73
    • 2 years ago
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    hope it was \(\frac{20}{56}\)

  37. jvaldez45
    • 2 years ago
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    yea

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