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

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

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

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

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

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

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

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
or how did i get the whole equation?
 one year 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.
 one year 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 \(a=1,b=a,c=a+80\)
 one year 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
 one year ago

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

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

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

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

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
hold on, you are not solving \((a20)(a+16)=0\) but rather \[(a20)(a+16)\geq 0\]
 one year 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 \(a\geq 20\)
 one year 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 \(a\geq 20\)
 one year 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?
 one year ago

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

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

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

jvaldez45 Group TitleBest ResponseYou've already chosen the best response.0
12 and 8
 one year ago

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

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

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