Question

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

Answer 1

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

Answer 2

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

Answer 3

are both real?

Answer 4

oh ok we can solve that using the quadratic formula

Answer 5

ok

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

how did you get that?

Answer 11

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

Answer 12

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

Answer 13

yes

Answer 14

or how did i get the whole equation?

Answer 15

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.

Answer 16

thanks

Answer 17

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 \(a=1,b=a,c=a+80\)

Answer 18

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

Answer 19

oh, what @phi said

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

a=-16

Answer 25

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

Answer 26

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 \(a\geq 20\)

Answer 27

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

Answer 28

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

Answer 29

not necessary, just look

Answer 30

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

Answer 31

you can eyeball the total length

Answer 32

12 and 8

Answer 33

right, for a total of 20

Answer 34

thanks just found the answer. thank you

Answer 35

yw

Answer 36

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

Answer 37

yea