## Destinyyyy one year ago Help..

1. anonymous

Solve by completing the square. x^2+11x-9=0

2. anonymous

@Nnesha

3. Michele_Laino

we can write this: $\Large {x^2} + 11x - 9 = {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9$

4. anonymous

Um where did you get 121 and 4 from???

5. Michele_Laino

since we have this: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}$

6. anonymous

Um okay.. Can you start at the beginning on how to solve this?

7. Michele_Laino

ok!

8. Michele_Laino

I rewrite the left side of your equation, adding and subtracting the same quantity, namely 121/4, so I get this equation: $\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0$

9. Michele_Laino

then, I use this identity: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}$

10. Michele_Laino

so I can rewrite your starting equation, as below: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0$

11. Michele_Laino

next, I nothe that: $\Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}$

12. Michele_Laino

after a substitution, I get this: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0$

13. Michele_Laino

or: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} = \frac{{157}}{4}$

14. anonymous

Uh? Im here- x^2+11x-9=0 (x+11/2)^2 -9=0 x^2+ 121/4 -9=0 I dont understand past that

15. anonymous

Ive been using this method or factoring https://www.youtube.com/watch?v=bclm1tJB-3g

16. Michele_Laino

In order to get an equivalent equation, I have to add and contemporarily I have to subtract 121/4, namely the same quantity

17. anonymous

Um okay

18. Michele_Laino

so I get this equation: $\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0$

19. Michele_Laino

the first three terms at the left side are the square of the subsequent binomial: $\Large {\left( {x + \frac{{11}}{2}} \right)^2}$

20. anonymous

Okay..

21. Michele_Laino

so, substituting, I can write this: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0$

22. anonymous

Okay

23. Michele_Laino

next I do this computation: $\Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}$

24. Michele_Laino

and, again I substitute: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0$

25. anonymous

What? I seriously do not understand. How in the world do you get 157? I understand that (x+ 11/2)^2 equals x^2 +121/4 .. I assume the 11 over 2 is 11x divided by 2 which cant happen so it stays as a fraction. You said something about subtracting 121/4 and I assumed you meant to the other side of the equal sign but wasn't entirely sure. Im really trying to follow what your saying and I'm good in math but all this makes absolutely no sense so far.

26. anonymous

Im still where I was earlier.

27. Michele_Laino

here is more details of my computation: |dw:1440704847919:dw|

28. Michele_Laino

here are*

29. anonymous

Where are you getting the negative in front of 121/4 from?

30. Michele_Laino

since I have used +121/4 in order to have the square of binomial, then the remaining -121/4 has to be summed to -9

31. anonymous

"then the remaining -121/4 has to be summed to -9" ??

32. Michele_Laino

yes! correct!

33. Nnesha

according to the video u watched 1st) move the constant term to the right side 2nd) divide b by 2 and then take square of the result in that case you you would add $\rm (\frac{ b }{ 2})^2$to the right side but if you keep the constant term at left side then you should subtract

34. anonymous

@Michele_Laino I put question marks at the end saying what to it... @Nnesha yeah I tried that and 11 divided by 2 is 5.5 so I stopped

35. Nnesha

36. Nnesha

that's right 5.5 which is same as 11/2 u got the decimal but he/she is keptt the fraction form but same thing! :=)

37. anonymous

Yes I understood that part. After that I get lost

38. Nnesha

so do you want to move the constant to the right side or do you want to keep it at left side ?

39. anonymous

The video says to move it to the right.. But I honestly dont know.

40. Nnesha

you will get the same answer

41. Michele_Laino

we have to add to both sides +157/4, so we get this: $\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} + \frac{{157}}{4} = \frac{{157}}{4}$

42. anonymous

I understand the adding to both sides but I dont get where you got 157 from

43. Nnesha

ye so doesn't matter you can move the constant term to right side at the beginning or after completing the square of (x^2+11x)

44. Nnesha

when you add -9 and -121/4 you will get that -157/4

45. Nnesha

$\huge\rm x^2+11x=9$ divide b by 2 and then take square of that result

46. anonymous

Do you mean (x^2+121/4) or 11/2= 5.5^2= 30.25

47. anonymous

Can you explain this more " when you add -9 and -121/4 you will get that -157/4 " because im not getting -157... -9+-121= -130 .. I dont see where the plus sign came from.

48. Nnesha

yes right $\huge\rm (x+5.5)^2=9+(5.5)^2$ okay i will

49. Nnesha

|dw:1440706398196:dw| there is one under 9 so 4 is the common denominator right ?

50. anonymous

Okay now I understand where the -157 come from.

51. anonymous

5.5*5.5= 30.25 +9= 39.25

52. Nnesha

yes right common denominator multiply the numerator of first fraction by the denominator of 2nd fraction multiply the numerator of 2nd fraction by the denominator of first fraction |dw:1440706515919:dw|

53. anonymous

Can you write it down how to solve this on paper and take a picture and post it on here? Maybe I can actually understand then

54. Nnesha

yes right 39.25 is same as 157/4

55. Nnesha

same thing basically

56. Nnesha

and after that i'm pretty sure you know how to solve for x right :D

57. anonymous

Yeah if I knew what the equation was.. All I have written down in front of me is x^2+11x+ ____=-9+____ x^2+11x-9=0 (x+11/2)^2 -9=0 x^2+ 121/4 -9=0

58. Nnesha

uh-oh why it's negative 9 at right sdie ?

59. Nnesha

side*

60. anonymous

Its suppose to be a positive mistype

61. Nnesha

yes right

62. anonymous

I have x^2- 157/4 =0

63. Nnesha

it's not x^2

64. Nnesha

cuz here |dw:1440706885310:dw| like this

65. Nnesha

or you can skip that factor step shortcut!|dw:1440707138935:dw|

66. Nnesha

divide b by 2 write that in the parentheses square would be to (x+b/2)^2 but you would add square of (b/2)^2 to the right side

67. Nnesha

that's another of saying x^2+!1x+ 30.25 = (x+5.5)^2 x+5.5 is a factor of x^2+!1x+30.25

68. anonymous

Yeah

69. Nnesha

ask question if you didn't get what i mean ......

70. anonymous

I understand that

71. Nnesha

this is good example here $\huge\rm x^2+bx+C=0$$\huge\rm x^\color{Red}{2}+Bx=-C$$\huge\rm (x+\frac{ b }{ 2})^\color{red}{2}=-C +(\frac{ b }{ 2 })^2$

72. anonymous

Okay.. I have (x+5.5)^2=+- 39.25 square root x+5.5= +- square root 39.25 x+5.5=+-

73. Nnesha

yes right $x+\frac{ 11 }{ 2 } = \pm \sqrt{ \frac{ {157} }{ 4 }}$ or you can keep it in fraction form

74. Nnesha

add you can see 4 is perfect square root so can take square of 4

75. anonymous

Okay I get it.. I think

76. Nnesha

if keep the decimal we will not be able to know if there is any perfect square root or not

77. anonymous

The final answer is x= 11/2 +- square root 157/ 4

78. Nnesha

okay practice 2 or 3 time you will definitely understand

79. Nnesha

hmmmm

80. anonymous

I hope so

81. anonymous

Opp negative in front of the 11

82. Nnesha

$x+\frac{ 11 }{ 2 } = \pm \sqrt{ \frac{ {157} }{ 4 }}$ $x+\frac{ 11 }{ 2 } =\frac{ \sqrt{157} }{ \sqrt{4}}$ square root of 4 is 2 right

83. Nnesha

$\huge\rm x=\frac{ -11 }{ 2 } \pm \frac{ \sqrt{157} }{ 2}$ so same denominator

84. anonymous

Ohh yeah

85. Nnesha

yes good luck!

86. anonymous

Thanks