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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
 2 years ago
 2 years ago
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
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.2
what is the probability that the roots of the equation what?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
what it the probability that the roots of the equation are real?
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
are both real?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
oh ok we can solve that using the quadratic formula
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
what would be the b value of the equation. is it a(x+1)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
\[x^2+ax+a+80=0 \] \[x=\frac{a\pm\sqrt{a^24a320}}{2}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
no if you use the quadratic formula, \(a=1,b=a, c=a+80\) different \(a\) of course
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
if these are to be real numbers, that means you must have \[a^24a300\geq 0\]
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
how did you get that?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
by some miracle this factors as \[(a20)(a+16)\geq 0\] so we can actually solve
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
how did i get \(a^24a320\)?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
or how did i get the whole equation?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
ok quadratic formula tells you the solution to \(ax^2+bx+c=0\) is \(x=\frac{b\pm\sqrt{b^24ac}}{2a}\) in your case \(a=1,b=a,c=a+80\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
when 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
oh, what @phi said
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
okay so after i find my values what do I do?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
your last job is to solve for \(a\) \[a^24a320\geq 0\]
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
i found my values and they are a=20 and a+16
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
this factors as \((a20)(a+16)\geq 0\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
hold on, you are not solving \((a20)(a+16)=0\) but rather \[(a20)(a+16)\geq 0\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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 \(a\geq 20\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
your final job is to find what portion of the interval \([28,28]\) satisfies \(a\leq 16\) or \(a\geq 20\)
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
okay so i would have to find the interval of the it by integrating it?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
not necessary, just look
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
favorable part is \([28,16]\cup [20,28]\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
you can eyeball the total length
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
right, for a total of 20
 2 years ago

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
thanks just found the answer. thank you
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
hope it was \(\frac{20}{56}\)
 2 years ago
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