anonymous
  • anonymous
Just for verification, can you please check my answer? Question: 3(x+2) = 6 ------ 2(x+1) my answer: 6(x^2+3x+1) Work: expanded first after multiplying both sides with 2(x+1).==> 6x^2+18+6==> took 6 out. to get final answer. Am i correct
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
What are your instructions?
anonymous
  • anonymous
Solve for x... shoot that means its incomplete.
anonymous
  • anonymous
Yep

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anonymous
  • anonymous
im stuck :( whats the next step? i cant think of any factors of 6x^2+18x+6... ah am i not suppose to factor, and use quadratic formula?
anonymous
  • anonymous
Hang on, what is the equation you just posted equal to?
anonymous
  • anonymous
my previous post is the multiplication of both sides with 2x+2
anonymous
  • anonymous
You have 6x^2+18x+6=0 right now, right?
anonymous
  • anonymous
like (3x+6)(2x+2) = 6 =>6x^2+18+12 = 6 ==> 6x^2+18+6
anonymous
  • anonymous
yes = 0
anonymous
  • anonymous
Perfect, so then you can factor out the 6 like you did before and use the quadratic formula on what's in the parentheses
anonymous
  • anonymous
ok one sec.
anonymous
  • anonymous
ok so it would be ___ ___ 3 ±√-1 and 3 ±√-1 ------ ------- 2 2
anonymous
  • anonymous
just the quadratic formula answers
anonymous
  • anonymous
*First one = + not ± *Second one = - not ±
anonymous
  • anonymous
Can you show me how you set up the quadratic formula because I got \[(-3\pm \sqrt{5})/2\] for the answer
anonymous
  • anonymous
umm ok -b ± √-b - 4ac --------------- 2a so -3±√3-4(1*1) / 2(1) * hey im sorry but can you help me in an hour? im at work and i over procrastinated and i need to finish my work, but ill be back online within an hour or hour and a half.
anonymous
  • anonymous
I gotta go soon, but FYI your formula's wrong: it's supposed to be b^2-4ac under the radical
anonymous
  • anonymous
Oh im sorry that was a typo...in both.
anonymous
  • anonymous
Ah i see, i did this within my previous answer too.
anonymous
  • anonymous
Does that solve the problem?
anonymous
  • anonymous
yes, but what would my final answer look like? 6 ((-3±√-5)/2) ?
anonymous
  • anonymous
No, because the 6 we factored out earlier doesn't affect the final answer because if we divided both sides by 6 before using the quadratic formula, we'd get x^2+3x+2=0 since 0/6=0
anonymous
  • anonymous
I hope that made some sense
anonymous
  • anonymous
So, your final answer would be \[(-3\pm \sqrt{-5})/2\]
anonymous
  • anonymous
Because the 6 canceled earlier
anonymous
  • anonymous
hmm ok i think i understand. because we divided 6, theres no need for its return. so i would just write x = (-3±√-5)/2
anonymous
  • anonymous
Yes! :)
anonymous
  • anonymous
Unless your book/teacher want you to write the two answers separately
anonymous
  • anonymous
In which case it would be \[x=(-3+ \sqrt{-5})/2\] and \[x=(-3-\sqrt{-5})/2\]
anonymous
  • anonymous
Any more questions?
anonymous
  • anonymous
Umm that's all for now, i should do my work now. But thank you very much :) I'll most likely post more questions within an hour or so to check my work so make sure i don't make mistakes like this again :p
anonymous
  • anonymous
Thanks again, bye
anonymous
  • anonymous
No prob. Bye

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