Help..

- Destinyyyy

Help..

- chestercat

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- Destinyyyy

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

- Destinyyyy

- 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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## More answers

- Destinyyyy

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

- Michele_Laino

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

- Destinyyyy

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

- Michele_Laino

ok!

- 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

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

- 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

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

- Michele_Laino

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

- Michele_Laino

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

- 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

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

- 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

Um okay

- Michele_Laino

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

- 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

Okay..

- Michele_Laino

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

- Destinyyyy

Okay

- Michele_Laino

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

- Michele_Laino

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

- 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

Im still where I was earlier.

- Michele_Laino

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

- Michele_Laino

here are*

- Destinyyyy

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

- 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

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

- Michele_Laino

yes! correct!

- 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

@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

http://prntscr.com/89mc1g
http://prntscr.com/89mbs3
so you need to be careful about this

- 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

Yes I understood that part. After that I get lost

- Nnesha

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

- Destinyyyy

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

- Nnesha

you will get the same answer

- 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

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

- 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

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

- Nnesha

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

- Destinyyyy

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

- 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

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

- Nnesha

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

- Destinyyyy

Okay now I understand where the -157 come from.

- Destinyyyy

5.5*5.5= 30.25 +9= 39.25

- 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

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

yes right 39.25 is same as 157/4

- Nnesha

same thing basically

- Nnesha

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

- 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

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

- Nnesha

side*

- Destinyyyy

Its suppose to be a positive mistype

- Nnesha

yes right

- Destinyyyy

I have x^2- 157/4 =0

- Nnesha

it's not x^2

- Nnesha

cuz here |dw:1440706885310:dw|
like this

- Nnesha

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

- 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

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

Yeah

- Nnesha

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

- Destinyyyy

I understand that

- 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

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

- Nnesha

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

- Nnesha

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

- Destinyyyy

Okay I get it.. I think

- Nnesha

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

- Destinyyyy

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

- Nnesha

okay practice 2 or 3 time you will definitely understand

- Nnesha

hmmmm

- Destinyyyy

I hope so

- Destinyyyy

Opp negative in front of the 11

- 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

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

- Destinyyyy

Ohh yeah

- Nnesha

yes good luck!

- Destinyyyy

Thanks

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