## anonymous 3 years ago Need some Assistance. (x-3)^2/(x+1)(x-2)>0 I'm supposed to solve it and express my answer in interval notation.

1. anonymous

|dw:1372837027496:dw|

2. zzr0ck3r

is this $\frac{(x-3)^{2}}{(x+1)(x-2)}$?

3. anonymous

Yes. that is the equation.

4. zzr0ck3r

ok so the top is >0 for all x does not equal 3 right?

5. anonymous

Right.

6. zzr0ck3r

so we need to make sure the bottom is > 0 right?

7. anonymous

Yes.

8. anonymous

|dw:1372837247856:dw|

9. anonymous

i have forgot 3

10. zzr0ck3r

the bottom is a parabola that opens up, what are its x intercepts?

11. anonymous

-1,2 Right?

12. zzr0ck3r

right

13. zzr0ck3r

so the bottom is greater than 0 when x<-1 and x>2, because it opens up and crosses y=0 at -1,2 with me?

14. anonymous

So far yes.

15. zzr0ck3r

so we know the top is grater than 0 when x is not 3 we know the bottom is greater than 0 when x<-1 and x>2 so the whole thing is greater than 0 when x<-1 and x>2 and x does not equal 3

16. zzr0ck3r

(-infinity,-1)U(2,3)U(3,infinity)

17. anonymous

Okay, I get it now.

18. zzr0ck3r

good deal:)

19. anonymous

I have another quick question

20. zzr0ck3r

oki

21. anonymous

I had another problem like this and I solved it and got x<-1. How would I write that in interval notation?

22. zzr0ck3r

(-infinity,-1)

23. zzr0ck3r

if it was <= (-infinity,-1]

24. anonymous

The original problem was $3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x$

25. zzr0ck3r

3-(9/4)x>5-(1/4)x -(8/4)x>5-3 2x<-2 x<-1 correct

26. anonymous

$x <-1$ I wrote the interval notation like so (-infinity,-1). But for some reason, my professor marked off points and circled interval notation.

27. zzr0ck3r

you sure it didn't say set builder?

28. anonymous

The question said Solve the inequalities and express your answer in Interval notation.

29. zzr0ck3r

he/she is wrong http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm look at non ending interval note: we have ) not ] because we have < not <=

30. zzr0ck3r

just go talk to your teacher, I make mistakes grading all the time....

31. anonymous

Okay, thank you!

32. zzr0ck3r

np

33. anonymous

One more equation?

34. anonymous

Inequality*

35. zzr0ck3r

sure

36. anonymous

$|\frac{x=1 }{ 4}|\ge4$

37. zzr0ck3r

is that plus or minus?

38. anonymous

I know your supposed to get two answers. I got one which was $x \ge15$

39. anonymous

40. anonymous

$|\frac{ x+1 }{4 }|$

41. zzr0ck3r

ok so this is strange and you may need to think about it a bit, but when you have |a|=b we get a=b and a=-b when you have |a|<b we get -b<x<b when you have |a|>b we get a<b and a>-b

42. zzr0ck3r

the very last and should be or

43. zzr0ck3r

*when you have |a|>b we get a<b or a>-b

44. anonymous

Okay, So if I want the other answer I set the 4 to a negative answer?

45. zzr0ck3r

so you have abs((x+1)/4)>=4 so (x+1)/4>=4 gives the solutions you already have

46. zzr0ck3r

not exactly

47. zzr0ck3r

now we do (x+1)/4<=-4 x+1<=-16 x<=-17

48. zzr0ck3r

set it negative and flip the sign

49. anonymous

Ah, that's what confused me.. Didn't know what to do with my sign

50. zzr0ck3r

you could have done -(x+1)/4<=4 but as a rule I just make the "non absolute value side" negative

51. anonymous

Good rule! With interval notation, how would I write this?

52. anonymous

I think its $(-\infty,-17]U(\infty,15]$

53. zzr0ck3r

(-infinity,-17]U[15,infinity)

54. zzr0ck3r

think of the number line, write it in the order it would fall on the number line

55. anonymous

Oh, I see.

56. zzr0ck3r

and note that (infinity,5] does not make since, because if x were to be in there then infinity < x < 5

57. zzr0ck3r

<=5

58. anonymous

You lost me there..

59. anonymous

But, Thank you so much for the help.