Destinyyyy
  • Destinyyyy
Help..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Destinyyyy
  • Destinyyyy
Solve by completing the square. x^2+11x-9=0
Destinyyyy
  • Destinyyyy
@Nnesha
Michele_Laino
  • Michele_Laino
we can write this: \[\Large {x^2} + 11x - 9 = {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9\]

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Destinyyyy
  • Destinyyyy
Um where did you get 121 and 4 from???
Michele_Laino
  • Michele_Laino
since we have this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}\]
Destinyyyy
  • Destinyyyy
Um okay.. Can you start at the beginning on how to solve this?
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
I rewrite the left side of your equation, adding and subtracting the same quantity, namely 121/4, so I get this equation: \[\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0\]
Michele_Laino
  • Michele_Laino
then, I use this identity: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = {x^2} + 11x + \frac{{121}}{4}\]
Michele_Laino
  • Michele_Laino
so I can rewrite your starting equation, as below: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0\]
Michele_Laino
  • Michele_Laino
next, I nothe that: \[ \Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}\]
Michele_Laino
  • Michele_Laino
after a substitution, I get this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0\]
Michele_Laino
  • Michele_Laino
or: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} = \frac{{157}}{4}\]
Destinyyyy
  • Destinyyyy
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
Destinyyyy
  • Destinyyyy
Ive been using this method or factoring https://www.youtube.com/watch?v=bclm1tJB-3g
Michele_Laino
  • 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
Destinyyyy
  • Destinyyyy
Um okay
Michele_Laino
  • Michele_Laino
so I get this equation: \[\Large {x^2} + 11x + \frac{{121}}{4} - \frac{{121}}{4} - 9 = 0\]
Michele_Laino
  • Michele_Laino
the first three terms at the left side are the square of the subsequent binomial: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2}\]
Destinyyyy
  • Destinyyyy
Okay..
Michele_Laino
  • Michele_Laino
so, substituting, I can write this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{121}}{4} - 9 = 0\]
Destinyyyy
  • Destinyyyy
Okay
Michele_Laino
  • Michele_Laino
next I do this computation: \[\Large - \frac{{121}}{4} - 9 = - \frac{{157}}{4}\]
Michele_Laino
  • Michele_Laino
and, again I substitute: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} = 0\]
Destinyyyy
  • Destinyyyy
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.
Destinyyyy
  • Destinyyyy
Im still where I was earlier.
Michele_Laino
  • Michele_Laino
here is more details of my computation: |dw:1440704847919:dw|
Michele_Laino
  • Michele_Laino
here are*
Destinyyyy
  • Destinyyyy
Where are you getting the negative in front of 121/4 from?
Michele_Laino
  • 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
Destinyyyy
  • Destinyyyy
"then the remaining -121/4 has to be summed to -9" ??
Michele_Laino
  • Michele_Laino
yes! correct!
Nnesha
  • 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
Destinyyyy
  • Destinyyyy
@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
Nnesha
  • Nnesha
http://prntscr.com/89mc1g http://prntscr.com/89mbs3 so you need to be careful about this
Nnesha
  • 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! :=)
Destinyyyy
  • Destinyyyy
Yes I understood that part. After that I get lost
Nnesha
  • Nnesha
so do you want to move the constant to the right side or do you want to keep it at left side ?
Destinyyyy
  • Destinyyyy
The video says to move it to the right.. But I honestly dont know.
Nnesha
  • Nnesha
you will get the same answer
Michele_Laino
  • Michele_Laino
we have to add to both sides +157/4, so we get this: \[\Large {\left( {x + \frac{{11}}{2}} \right)^2} - \frac{{157}}{4} + \frac{{157}}{4} = \frac{{157}}{4}\]
Destinyyyy
  • Destinyyyy
I understand the adding to both sides but I dont get where you got 157 from
Nnesha
  • 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)
Nnesha
  • Nnesha
when you add -9 and -121/4 you will get that -157/4
Nnesha
  • Nnesha
\[\huge\rm x^2+11x=9\] divide b by 2 and then take square of that result
Destinyyyy
  • Destinyyyy
Do you mean (x^2+121/4) or 11/2= 5.5^2= 30.25
Destinyyyy
  • Destinyyyy
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.
Nnesha
  • Nnesha
yes right \[\huge\rm (x+5.5)^2=9+(5.5)^2\] okay i will
Nnesha
  • Nnesha
|dw:1440706398196:dw| there is one under 9 so 4 is the common denominator right ?
Destinyyyy
  • Destinyyyy
Okay now I understand where the -157 come from.
Destinyyyy
  • Destinyyyy
5.5*5.5= 30.25 +9= 39.25
Nnesha
  • 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|
Destinyyyy
  • Destinyyyy
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
Nnesha
  • Nnesha
yes right 39.25 is same as 157/4
Nnesha
  • Nnesha
same thing basically
Nnesha
  • Nnesha
and after that i'm pretty sure you know how to solve for x right :D
Destinyyyy
  • Destinyyyy
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
Nnesha
  • Nnesha
uh-oh why it's negative 9 at right sdie ?
Nnesha
  • Nnesha
side*
Destinyyyy
  • Destinyyyy
Its suppose to be a positive mistype
Nnesha
  • Nnesha
yes right
Destinyyyy
  • Destinyyyy
I have x^2- 157/4 =0
Nnesha
  • Nnesha
it's not x^2
Nnesha
  • Nnesha
cuz here |dw:1440706885310:dw| like this
Nnesha
  • Nnesha
or you can skip that factor step shortcut!|dw:1440707138935:dw|
Nnesha
  • 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
Nnesha
  • 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
Destinyyyy
  • Destinyyyy
Yeah
Nnesha
  • Nnesha
ask question if you didn't get what i mean ......
Destinyyyy
  • Destinyyyy
I understand that
Nnesha
  • Nnesha
this is good example here \[\huge\rm x^2+bx+C=0\]\[\huge\rm x^\color{Red}{2}+Bx=-C\]\[\huge\rm (x+\frac{ b }{ 2})^\color{red}{2}=-C +(\frac{ b }{ 2 })^2\]
Destinyyyy
  • Destinyyyy
Okay.. I have (x+5.5)^2=+- 39.25 square root x+5.5= +- square root 39.25 x+5.5=+-
Nnesha
  • Nnesha
yes right \[x+\frac{ 11 }{ 2 } = \pm \sqrt{ \frac{ {157} }{ 4 }}\] or you can keep it in fraction form
Nnesha
  • Nnesha
add you can see 4 is perfect square root so can take square of 4
Destinyyyy
  • Destinyyyy
Okay I get it.. I think
Nnesha
  • Nnesha
if keep the decimal we will not be able to know if there is any perfect square root or not
Destinyyyy
  • Destinyyyy
The final answer is x= 11/2 +- square root 157/ 4
Nnesha
  • Nnesha
okay practice 2 or 3 time you will definitely understand
Nnesha
  • Nnesha
hmmmm
Destinyyyy
  • Destinyyyy
I hope so
Destinyyyy
  • Destinyyyy
Opp negative in front of the 11
Nnesha
  • 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
Nnesha
  • Nnesha
\[\huge\rm x=\frac{ -11 }{ 2 } \pm \frac{ \sqrt{157} }{ 2}\] so same denominator
Destinyyyy
  • Destinyyyy
Ohh yeah
Nnesha
  • Nnesha
yes good luck!
Destinyyyy
  • Destinyyyy
Thanks

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