## jvaldez45 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. satellite73

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

2. satellite73

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

3. jvaldez45

are both real?

4. satellite73

oh ok we can solve that using the quadratic formula

5. jvaldez45

ok

6. jvaldez45

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

7. satellite73

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

8. satellite73

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

9. satellite73

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

10. jvaldez45

how did you get that?

11. satellite73

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

12. satellite73

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

13. jvaldez45

yes

14. satellite73

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. jvaldez45

thanks

17. satellite73

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. satellite73

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. satellite73

oh, what @phi said

20. jvaldez45

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

21. satellite73

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

22. jvaldez45

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

23. satellite73

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

24. jvaldez45

a=-16

25. satellite73

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

26. satellite73

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. satellite73

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

28. jvaldez45

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

29. satellite73

not necessary, just look

30. satellite73

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

31. satellite73

you can eyeball the total length

32. jvaldez45

12 and 8

33. satellite73

right, for a total of 20

34. jvaldez45

thanks just found the answer. thank you

35. satellite73

yw

36. satellite73

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

37. jvaldez45

yea