## anonymous 3 years ago If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0

1. anonymous

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

2. anonymous

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

3. anonymous

are both real?

4. anonymous

oh ok we can solve that using the quadratic formula

5. anonymous

ok

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

how did you get that?

11. anonymous

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

12. anonymous

how did i get $$a^2-4a-320$$?

13. anonymous

yes

14. anonymous

or how did i get the whole equation?

15. 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.

16. anonymous

thanks

17. 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 $$a=1,b=a,c=a+80$$

18. 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

19. anonymous

oh, what @phi said

20. anonymous

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

21. anonymous

your last job is to solve for $$a$$ $a^2-4a-320\geq 0$

22. anonymous

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

23. anonymous

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

24. anonymous

a=-16

25. anonymous

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

26. 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 $$a\geq 20$$

27. anonymous

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

28. anonymous

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

29. anonymous

not necessary, just look

30. anonymous

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

31. anonymous

you can eyeball the total length

32. anonymous

12 and 8

33. anonymous

right, for a total of 20

34. anonymous

thanks just found the answer. thank you

35. anonymous

yw

36. anonymous

hope it was $$\frac{20}{56}$$

37. anonymous

yea