## anonymous one year ago can anyone help??

1. anonymous

2. anonymous

3. anonymous

i think i just neeed to do the last 4 steps its asking at the bottom

4. anonymous

can anyone help or understand how to do that?

5. xapproachesinfinity

what you think u should be?

6. xapproachesinfinity

any ideas?

7. xapproachesinfinity

oh there two pages last 4 you are reffering to page 2 right?

8. anonymous

its all one problem

9. xapproachesinfinity

okay! first pages looks like an example to me

10. anonymous

i think i need to answer the 4 questions at the bottom of pg 2

11. xapproachesinfinity

anyways we are doing this $$\sqrt{(x^2-x)+7}=2(x^2-x)-1$$ 1) is about the u

12. xapproachesinfinity

any idea?

13. anonymous

there is no u there, but in pg 11 it says let u= t^2-3t is that right?

14. xapproachesinfinity

it say what would u represent i may reading this incorrectly lol

15. xapproachesinfinity

they want you to suggest a u for that equation

16. anonymous

idk im very confused i dont see a U?? :S

17. xapproachesinfinity

no there is no u in the problem! what they are telling you is do substitution instead of solving for x we solve for a new variable u but we need first to make some kind of substitution if you looked at the first page you have they give you an example

18. xapproachesinfinity

i think you didn;t read what that page is saying otherwise you would be confused what u is lol

19. xapproachesinfinity

because that first page explains it all!

20. anonymous

i did read it.

21. xapproachesinfinity

lol! okay no problem so we need to do substitution to some variable u I'm speaking of this equation now $$\sqrt{(x^2-x)+7}=2(x^2-x)-1$$

22. xapproachesinfinity

this equation looks a mess for now that's why we do u to have only one good looking variable

23. anonymous

so we need to replace (x^2-x) with u right

24. xapproachesinfinity

so we can solve easily ah yes you are catching good going so far

25. xapproachesinfinity

$$\sqrt{u+7}=2u-1$$ letting u=x^2-x yes

26. xapproachesinfinity

now we can solve for u before we go back to x

27. xapproachesinfinity

following?

28. anonymous

yea

29. xapproachesinfinity

$$\sqrt{u+7}=2u-1 \Longrightarrow u+7=(2u-1)^2$$

30. xapproachesinfinity

agree with this step or no?

31. xapproachesinfinity

i took square of both sides

32. anonymous

to get rid of the sqrt, yea

33. xapproachesinfinity

good! so now $$u+7=4u^2-4u+1 \Longrightarrow 4u^2-5u-6=0$$

34. xapproachesinfinity

we got a quadratic!

35. xapproachesinfinity

can we factor that!

36. anonymous

wait

37. xapproachesinfinity

yes! any questions

38. anonymous

to get 4u^2 -5u-6, you just did -u-7 from both sides right?

39. xapproachesinfinity

yes!

40. anonymous

and then we need to factr that to (4u+3)(u-2)?

41. xapproachesinfinity

now we factor to $$(u-2)(4u+3)=0$$

42. xapproachesinfinity

yes excellent

43. anonymous

and use zpp?

44. xapproachesinfinity

solve for u what is zpp lol never heard this haha

45. anonymous

zero product property lol

46. xapproachesinfinity

i guess you mean u-2=0 or 4u+3=0 u=2 or u=-3/4

47. anonymous

yep

48. xapproachesinfinity

eh i see i never named anyhing it just makes sense to me with no name haha

49. xapproachesinfinity

now we solved for u but we still not finished for x

50. xapproachesinfinity

remember that we let u=x^2-x right

51. anonymous

so for the first question its u= *3/4 and u=2 right

52. xapproachesinfinity

so we have $$x^2-x=2 ~~or~~ x^2-x=-3/4$$

53. xapproachesinfinity

oh no first question to just answer u=x^2-x that is it

54. xapproachesinfinity

we are actually doing all those 4 question in one go

55. anonymous

ohh okay

56. xapproachesinfinity

once we finish you put it in order 1 put what we did 2 put what we did

57. xapproachesinfinity

here i went to answer those as one question lol they supposed to be one it is just that they want you to understand it part by part

58. anonymous

okay

59. xapproachesinfinity

2 we did we wrote the equATION WITH ONLY U

60. anonymous

gotcha

61. xapproachesinfinity

3 we did we solved for u u=2 u=-3/4

62. xapproachesinfinity

we are in 4 now :)

63. xapproachesinfinity

$$x^2-x=2 ~~or x^2-x=-3/4$$ we solve this now

64. xapproachesinfinity

65. xapproachesinfinity

$$x^2-x-2=0 ~~~or ~~~ 4x^2-4x+3=0$$

66. xapproachesinfinity

looks good to you?

67. anonymous

no lol

68. xapproachesinfinity

may be i wen too fast haha x^2-x=2 x^2-x-2=0 after subtracting 2 from both ssides

69. anonymous

how did you get 4x^2-4x+3

70. xapproachesinfinity

the other one x^2-x=-3/4 ===> 4(x^2-x)=-3 (multiplied 4 both sides) 4x^2-4x=-3 4x^2-4x+3=0 added 3 to both sides

71. xapproachesinfinity

clear now i guess haha

72. xapproachesinfinity

i just wanted to have a nice format i dislike fractions haha

73. xapproachesinfinity

good or nor good?

74. anonymous

i was just trying to understand it sorry, im a slow learner lol i got it tho

75. anonymous

i got it

76. xapproachesinfinity

hmm you are good learner as i see it :)

77. anonymous

aw thanks aha :))

78. xapproachesinfinity

we solve $$x^2-x-2=0 \Longrightarrow (x+1)(x-2)=0$$

79. xapproachesinfinity

this one 4x^2-4x+3=0 does not have real solutions

80. xapproachesinfinity

so we just solve (x+1)(x-2)=0

81. xapproachesinfinity

zpp lol

82. anonymous

x=1 and 2

83. anonymous

-1,+2

84. xapproachesinfinity

yes great job

85. xapproachesinfinity

that for step 4 and we are good!

86. anonymous

cool thanks so much

87. xapproachesinfinity

no problem

88. anonymous

you are a great teacher aha

89. xapproachesinfinity

not really :) i suck at some stuff haha