## jvaldez45 Group Title If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0 one year ago one year ago

1. satellite73 Group Title

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

2. satellite73 Group Title

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

3. jvaldez45 Group Title

are both real?

4. satellite73 Group Title

oh ok we can solve that using the quadratic formula

5. jvaldez45 Group Title

ok

6. jvaldez45 Group Title

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

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. satellite73 Group Title

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

10. jvaldez45 Group Title

how did you get that?

11. satellite73 Group Title

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

12. satellite73 Group Title

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

13. jvaldez45 Group Title

yes

14. satellite73 Group Title

or how did i get the whole equation?

15. phi Group Title

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 Group Title

thanks

17. satellite73 Group Title

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 Group Title

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 Group Title

oh, what @phi said

20. jvaldez45 Group Title

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

21. satellite73 Group Title

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

22. jvaldez45 Group Title

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

23. satellite73 Group Title

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

24. jvaldez45 Group Title

a=-16

25. satellite73 Group Title

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

26. satellite73 Group Title

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 Group Title

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

28. jvaldez45 Group Title

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

29. satellite73 Group Title

not necessary, just look

30. satellite73 Group Title

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

31. satellite73 Group Title

you can eyeball the total length

32. jvaldez45 Group Title

12 and 8

33. satellite73 Group Title

right, for a total of 20

34. jvaldez45 Group Title

thanks just found the answer. thank you

35. satellite73 Group Title

yw

36. satellite73 Group Title

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

37. jvaldez45 Group Title

yea