anonymous
  • anonymous
complete the square: 3x^2-6x=24
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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eSpeX
  • eSpeX
Where are you stuck? If you are wondering how to start, you want to divide out your leading coefficient on your x^2 term.
anonymous
  • anonymous
you subtract 24 from both side, to zero out the equation, then you use the square this to solve
anonymous
  • anonymous
|dw:1370393853970:dw|

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anonymous
  • anonymous
okay i did the (b/2)^2 thing and then got 3x^2 - 6x +9 = 33. then i divided the 3 which resulted to 3(x^2-2x+3)=33 now i cant factor that, i think i did something wrong, i just dont know what it is
anonymous
  • anonymous
|dw:1370393983063:dw| just factor this out
eSpeX
  • eSpeX
You need to first divide out the leading coefficient \(3x^2-6x=24 \rightarrow x^2-2x=8\) Then take half of your second term, square it and then add that to both sides.
anonymous
  • anonymous
@eSpeX no they need to zero out the equation and then factor it
eSpeX
  • eSpeX
@thefukintoilet there is if you are following the instructions to "complete the square" in order to solve the equation.
anonymous
  • anonymous
oh ok im sorry @eSpeX
eSpeX
  • eSpeX
No worries, you never know where the student is in their lesson so I always try to adhere to the instructions offered.
eSpeX
  • eSpeX
Once you divide 3 out of every term in the equation it is gone and you no longer have to worry about it.
anonymous
  • anonymous
ohhhh okay thankss
eSpeX
  • eSpeX
You're welcome, do you know how to finish this problem then?
anonymous
  • anonymous
i got 10
eSpeX
  • eSpeX
hmm, I got 4, can you show me how you got your answer?
anonymous
  • anonymous
oh lol
anonymous
  • anonymous
3x^2-6x=24 x^2-2x=8 (-2/2) = -1^2 = 1 x^2-2x+1=8+1 (x-1)(x-1)=9 x-1=9 +1=+1 x=10
eSpeX
  • eSpeX
You need to add a step. After (x-1)(x-1)=9, you should have \((x-1)^2=9\) Then take the sqrt of both sides and solve for 'x'.
anonymous
  • anonymous
ohhh got it.
anonymous
  • anonymous
now i got 4, thaaank you so much.
eSpeX
  • eSpeX
You are welcome.

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