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anonymous
 3 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
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is the probability that the roots of the equation what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what it the probability that the roots of the equation are real?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok we can solve that using the quadratic formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what would be the b value of the equation. is it a(x+1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2+ax+a+80=0 \] \[x=\frac{a\pm\sqrt{a^24a320}}{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no if you use the quadratic formula, \(a=1,b=a, c=a+80\) different \(a\) of course

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if these are to be real numbers, that means you must have \[a^24a300\geq 0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by some miracle this factors as \[(a20)(a+16)\geq 0\] so we can actually solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did i get \(a^24a320\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or how did i get the whole equation?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0how 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok quadratic formula tells you the solution to \(ax^2+bx+c=0\) is \(x=\frac{b\pm\sqrt{b^24ac}}{2a}\) in your case \(a=1,b=a,c=a+80\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when you compute you get \[x=\frac{a\pm\sqrt{a^24a320}}{2}\] in your case so for these to be real, the discriminant \(a^24a320\) must be greater than or equal to zero, otherwise you have a negative number under the radical

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so after i find my values what do I do?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your last job is to solve for \(a\) \[a^24a320\geq 0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i found my values and they are a=20 and a+16

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this factors as \((a20)(a+16)\geq 0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold on, you are not solving \((a20)(a+16)=0\) but rather \[(a20)(a+16)\geq 0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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 \(a\geq 20\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your final job is to find what portion of the interval \([28,28]\) satisfies \(a\leq 16\) or \(a\geq 20\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so i would have to find the interval of the it by integrating it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not necessary, just look

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0favorable part is \([28,16]\cup [20,28]\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can eyeball the total length

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, for a total of 20

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks just found the answer. thank you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hope it was \(\frac{20}{56}\)
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