If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0

- anonymous

- jamiebookeater

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- anonymous

what is the probability that the roots of the equation what?

- anonymous

what it the probability that the roots of the equation are real?

- anonymous

are both real?

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## More answers

- anonymous

oh ok we can solve that using the quadratic formula

- anonymous

ok

- anonymous

what would be the b value of the equation. is it a(x+1)

- anonymous

\[x^2+ax+a+80=0 \]
\[x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}\]

- anonymous

no if you use the quadratic formula, \(a=1,b=a, c=a+80\)
different \(a\) of course

- anonymous

if these are to be real numbers, that means you must have
\[a^2-4a-300\geq 0\]

- anonymous

how did you get that?

- anonymous

by some miracle this factors as
\[(a-20)(a+16)\geq 0\] so we can actually solve

- anonymous

how did i get \(a^2-4a-320\)?

- anonymous

yes

- anonymous

or how did i get the whole equation?

- 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

thanks

- 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

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

oh, what @phi said

- anonymous

okay so after i find my values what do I do?

- anonymous

your last job is to solve for \(a\)
\[a^2-4a-320\geq 0\]

- anonymous

i found my values and they are a=20 and a+-16

- anonymous

this factors as \((a-20)(a+16)\geq 0\)

- anonymous

a=-16

- anonymous

hold on, you are not solving \((a-20)(a+16)=0\) but rather
\[(a-20)(a+16)\geq 0\]

- 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

your final job is to find what portion of the interval \([-28,28]\) satisfies \(a\leq -16\) or \(a\geq 20\)

- anonymous

okay so i would have to find the interval of the it by integrating it?

- anonymous

not necessary, just look

- anonymous

favorable part is \([-28,-16]\cup [20,28]\)

- anonymous

you can eyeball the total length

- anonymous

12 and 8

- anonymous

right, for a total of 20

- anonymous

thanks just found the answer. thank you

- anonymous

yw

- anonymous

hope it was \(\frac{20}{56}\)

- anonymous

yea

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