anonymous
  • anonymous
If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what is the probability that the roots of the equation what?
anonymous
  • anonymous
what it the probability that the roots of the equation are real?
anonymous
  • anonymous
are both real?

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anonymous
  • anonymous
oh ok we can solve that using the quadratic formula
anonymous
  • anonymous
ok
anonymous
  • anonymous
what would be the b value of the equation. is it a(x+1)
anonymous
  • anonymous
\[x^2+ax+a+80=0 \] \[x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}\]
anonymous
  • anonymous
no if you use the quadratic formula, \(a=1,b=a, c=a+80\) different \(a\) of course
anonymous
  • anonymous
if these are to be real numbers, that means you must have \[a^2-4a-300\geq 0\]
anonymous
  • anonymous
how did you get that?
anonymous
  • anonymous
by some miracle this factors as \[(a-20)(a+16)\geq 0\] so we can actually solve
anonymous
  • anonymous
how did i get \(a^2-4a-320\)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
or how did i get the whole equation?
phi
  • phi
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.
anonymous
  • anonymous
thanks
anonymous
  • anonymous
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 \(a=1,b=a,c=a+80\)
anonymous
  • anonymous
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
anonymous
  • anonymous
oh, what @phi said
anonymous
  • anonymous
okay so after i find my values what do I do?
anonymous
  • anonymous
your last job is to solve for \(a\) \[a^2-4a-320\geq 0\]
anonymous
  • anonymous
i found my values and they are a=20 and a+-16
anonymous
  • anonymous
this factors as \((a-20)(a+16)\geq 0\)
anonymous
  • anonymous
a=-16
anonymous
  • anonymous
hold on, you are not solving \((a-20)(a+16)=0\) but rather \[(a-20)(a+16)\geq 0\]
anonymous
  • anonymous
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 \(a\geq 20\)
anonymous
  • anonymous
your final job is to find what portion of the interval \([-28,28]\) satisfies \(a\leq -16\) or \(a\geq 20\)
anonymous
  • anonymous
okay so i would have to find the interval of the it by integrating it?
anonymous
  • anonymous
not necessary, just look
anonymous
  • anonymous
favorable part is \([-28,-16]\cup [20,28]\)
anonymous
  • anonymous
you can eyeball the total length
anonymous
  • anonymous
12 and 8
anonymous
  • anonymous
right, for a total of 20
anonymous
  • anonymous
thanks just found the answer. thank you
anonymous
  • anonymous
yw
anonymous
  • anonymous
hope it was \(\frac{20}{56}\)
anonymous
  • anonymous
yea

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