jvaldez45
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
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what is the probability that the roots of the equation what?
anonymous
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what it the probability that the roots of the equation are real?
jvaldez45
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are both real?
anonymous
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oh ok we can solve that using the quadratic formula
jvaldez45
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ok
jvaldez45
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what would be the b value of the equation. is it a(x+1)
anonymous
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\[x^2+ax+a+80=0 \]
\[x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}\]
anonymous
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no if you use the quadratic formula, \(a=1,b=a, c=a+80\)
different \(a\) of course
anonymous
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if these are to be real numbers, that means you must have
\[a^2-4a-300\geq 0\]
jvaldez45
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how did you get that?
anonymous
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by some miracle this factors as
\[(a-20)(a+16)\geq 0\] so we can actually solve
anonymous
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how did i get \(a^2-4a-320\)?
jvaldez45
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yes
anonymous
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or how did i get the whole equation?
phi
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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.
jvaldez45
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thanks
anonymous
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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 \(a=1,b=a,c=a+80\)
anonymous
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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
anonymous
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oh, what @phi said
jvaldez45
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okay so after i find my values what do I do?
anonymous
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your last job is to solve for \(a\)
\[a^2-4a-320\geq 0\]
jvaldez45
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i found my values and they are a=20 and a+-16
anonymous
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this factors as \((a-20)(a+16)\geq 0\)
jvaldez45
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a=-16
anonymous
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hold on, you are not solving \((a-20)(a+16)=0\) but rather
\[(a-20)(a+16)\geq 0\]
anonymous
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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 \(a\geq 20\)
anonymous
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your final job is to find what portion of the interval \([-28,28]\) satisfies \(a\leq -16\) or \(a\geq 20\)
jvaldez45
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okay so i would have to find the interval of the it by integrating it?
anonymous
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not necessary, just look
anonymous
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favorable part is \([-28,-16]\cup [20,28]\)
anonymous
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you can eyeball the total length
jvaldez45
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12 and 8
anonymous
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right, for a total of 20
jvaldez45
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thanks just found the answer. thank you
anonymous
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yw
anonymous
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hope it was \(\frac{20}{56}\)
jvaldez45
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yea