anonymous
  • anonymous
Solve: x2 + 2x - 1 = 0
Mathematics
katieb
  • katieb
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Nnesha
  • Nnesha
i'm pretty sure you know how to factor that :-
ParthKohli
  • ParthKohli
That can't be factored, honestly.
anonymous
  • anonymous
i got \[-2\pm \sqrt{8}> divided by 2\]

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anonymous
  • anonymous
oh wait
ParthKohli
  • ParthKohli
Looks alright to me. You can write that as \(-1 \pm \sqrt 2\)
Nnesha
  • Nnesha
factor 8
anonymous
  • anonymous
square root of 8 is a long decimal
anonymous
  • anonymous
\[x^2+2x=1,add~1 ~\to~both~sides\]
ParthKohli
  • ParthKohli
\[\sqrt 8 = 2 \sqrt 2\]
Nnesha
  • Nnesha
\[\huge\rm \frac{ -2 \pm \sqrt{4 \times 2}}{ 2}\] yep right that's why you to factor 8 4 times 2 =8 now take square root of 4
Nnesha
  • Nnesha
have**
anonymous
  • anonymous
thats what i got, but what next?
ParthKohli
  • ParthKohli
Then you've solved it.
anonymous
  • anonymous
1 + √2 and -1 - √2 -1 + √2 and -1 - √2 -1 + √3 and -1 + √2 -1 - √2 and 1 + √2
Nnesha
  • Nnesha
\[\huge\rm \frac{ -2 \pm 2\sqrt{ \times 2}}{ 2}\] 2 is common so take it out
anonymous
  • anonymous
oh
anonymous
  • anonymous
so,..|dw:1433904972339:dw|
anonymous
  • anonymous
i will fan all of u if you help me with 3 questions
anonymous
  • anonymous
is that it?
anonymous
  • anonymous
\[x^2+2x+1=2,(x+1)^2=2,x+1=\pm \sqrt{2}\] \[x=-1\pm \sqrt{2}\]
anonymous
  • anonymous
so i am right
Nnesha
  • Nnesha
\[\large\rm \frac{ \cancel 2(-1 \pm \sqrt{ 2})}{ \cancel2}\] yes you got it right!
triciaal
  • triciaal
|dw:1433908590016:dw|

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