## moongazer 3 years ago how do you get the domain of this:

1. moongazer
2. moongazer

how do you find it? step-by-step please :)

3. ganeshie8
4. moongazer

√(x^2 - 9x +20)/√(x^2 - 5x + 6) I already know the domain. But, how do you get it?

5. ganeshie8

first workout contraints for squareroot : the thing inside squareroot cannot be negative

6. ganeshie8

next work out constraints for denominator : denominator cannot be equal to 0

7. ganeshie8

consider numerator first : $$\sqrt{x^2 - 9x +20}$$

8. ganeshie8

can u work out domain for that ?

9. moongazer

{x|x>=5} is that right?

10. ganeshie8

somewhat right :) lets see how to solve

11. ganeshie8

x^2-9x+20 >= 0 x^2-4x-5x+20 >=0 x(x-4) -5(x-4) >= 0 (x-4)(x-5) >= 0 x <= 4 or x >= 5

12. ganeshie8

thats one constraint for numerator, lets find out the constraints for denominator

13. moongazer

ohhhh

14. moongazer

I think I am starting to understand it :)

15. ganeshie8

good :) since, denominator is also under radical, it must be >=0 : x^2 - 5x + 6 >= 0 x^2-3x-2x + 6 >= 0 x(x-3) -2(x-3) >= 0 (x-2)(x-3) >= 0 x <= 2 or x >= 3

16. ganeshie8

with those two things, we finished with the first step of finding constraints for radicals

17. ganeshie8

so far we have this : x <= 4 or x >= 5 x <= 2 or x >= 3

18. moongazer

yup, I understand it :)

19. ganeshie8

next one is denominator can never equal to 0. so, x^2 - 5x + 6 $$\ne$$ 0 x^2-3x-2x + 6 $$\ne$$ 0 x(x-3) -2(x-3) $$\ne$$ 0 (x-2)(x-3) $$\ne$$ 0 x $$\ne$$ 2 or x $$\ne$$ 3

20. moongazer

w8 Isn't it that it should be: x >= 4 or x >= 5 x >= 2 or x >= 3 ???

21. ganeshie8

lets combine all constraints and make a meaningful constraint

22. ganeshie8

which one ?

23. moongazer

(x-4)(x-5) >= 0 (x-2)(x-3) >= 0

24. ganeshie8

oh i got ur question, il give quick explanation

25. ganeshie8

you comfortable with parabola graph ?

26. moongazer

I think so

27. ganeshie8

il show u in graph why its x <=4 , x >=5

28. moongazer

I know few things about parabola

29. moongazer

ok

30. ganeshie8

im sketching this parabola : (x-4)(x-5) >= 0

31. ganeshie8

it intersect x axis at 4, 5 right ?

32. ganeshie8

|dw:1348496600444:dw|

33. ganeshie8

between 4 and 5, it is sinking down into x axis eh ?

34. ganeshie8

its becoming NEGATIVE

35. ganeshie8

|dw:1348496705657:dw|

36. moongazer

why did it intersect in x axis at 4, 5?

37. ganeshie8

when u factored you got : (x-4)(x-5)

38. ganeshie8

that means it intersects x-axis at 4 and 5

39. moongazer

ohh, ok what if it says (x-4)(x+5) does it intersect at 4 and -5 ?

40. ganeshie8

thats right, when it says, (x-4)(x+5) = 0 then it intersects at 4 and -5

41. ganeshie8

cuz for it to equal to 0, one of both of the factors must equal to 0

42. moongazer

|dw:1348498226449:dw|

43. moongazer

so the graph is like that?

44. ganeshie8

perfect, thats (x-4)(x+5)

45. moongazer

and the domain is x>=4,x<=-5 is it correct?

46. ganeshie8

thats right ! you became expert ;p

47. moongazer

for (x-4)(x+5)

48. moongazer

Thanks! I know understand it. :) let's now continue with our previous discussion :)

49. ganeshie8

great ! so far what we have

50. moongazer

next one is denominator can never equal to 0. so, x^2 - 5x + 6 ≠ 0 x^2-3x-2x + 6 ≠ 0 x(x-3) -2(x-3) ≠ 0 (x-2)(x-3) ≠ 0 x ≠ 2 or x ≠ 3

51. ganeshie8

yeah we almost done

52. ganeshie8

lets put down all the 3 constraints we got in one place :

53. ganeshie8

1) numerator radical : x <= 4 or x >= 5 2) denominator radical : x <= 2 or x >= 3 3) denominator cant be 0 : x ≠ 2 or x ≠ 3

54. moongazer

I understand it. What's next?

55. ganeshie8

see 2) and 3)

56. ganeshie8

can we combine them as one, like this : 2) 3) : x < 2 or x > 3

57. moongazer

I'll be back just continue.

58. moongazer

someone is calling me

59. ganeshie8

ok after combining 2) and 3) we are left with : 1) numerator radical : x <= 4 or x >= 5 2) x < 2 or x > 3 this is the last step. see if u can figure out

60. ganeshie8

out of x < 2 & x <= 4, x < 2 is more restrictive, so we have this in our domain : $$\color{green}{x < 2}$$

61. ganeshie8

out of x > 3 & x <= 4 x > 3 , x <= 4 , which is same as : $$\color{green}{3<x\le 4}$$

62. ganeshie8

and of course the last one, $$\color{green}{x >= 5}$$

63. moongazer

sorry for being away, I'm here again :)

64. ganeshie8

np :) see if above stuff makes sense

65. moongazer

it makes sense but why in the final answer they used "or" not "and" "," "then" or anything else dom K: {x|x<2 or 3<x<=4 or x>=5}

66. ganeshie8

yeah we always combine domain with "or" why ? cuz its meaningful. let me ask u a q

67. ganeshie8

suppose we write : x < 2 and x >=5

68. moongazer

for the domain it is ALWAYS "or" ?

69. ganeshie8

is that possible ? can x be BOTH less than 2 and greater than 5 at the SAME time ha ?

70. ganeshie8

yes always "or"

71. moongazer

yes, it is not possible

72. ganeshie8

is "1" less than 2 and greater than "5" ?

73. ganeshie8

yeah, but if we put it like this : either "1" is less than 2 or greater than "5"

74. ganeshie8

how meaningful it is, so while phrasing domain also, we use commonsense aswell :)

75. moongazer

yes, Thank you very much for helping me :)

76. ganeshie8

np i learned as well. i forgot these long back in college... i enjoyed going thru these again.. thnks you too :)