anonymous
  • anonymous
3x^2-x-4=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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misty1212
  • misty1212
they really got you factoring, don't they?
anonymous
  • anonymous
yes i took algebra 3 and this is all we are doing is factoring
misty1212
  • misty1212
algebra 3? i didnt know they had that many algebras?

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More answers

anonymous
  • anonymous
neither did i until i went to this school
misty1212
  • misty1212
try \[(3x-4)(x+1)=0\]
arindameducationusc
  • arindameducationusc
You can use quadratic equation also.....
misty1212
  • misty1212
sure but that is more complicated, my guess is since one factors, they all factor
arindameducationusc
  • arindameducationusc
I agree with Misty's factorisation...... that's more easy....
misty1212
  • misty1212
do you know what to do once you have \[(3x-4)(x+1)=0\]?
anonymous
  • anonymous
it worked thank you @misty1212
misty1212
  • misty1212
\[\color\magenta\heartsuit\]
anonymous
  • anonymous
i have one more problem if you don't mind helping :)
misty1212
  • misty1212
not at all happy to
anonymous
  • anonymous
\[6x^2-13x+6=0\]
misty1212
  • misty1212
bet we can factor this one too huh?
anonymous
  • anonymous
yeah pretty sure . my problem is trying to figure out what plus what would equal them middle number , and that would also equal the last number if multiplied
misty1212
  • misty1212
\[(2 x-3) (3 x-2) = 0\] seems to work
arindameducationusc
  • arindameducationusc
arindameducationusc
  • arindameducationusc
use this b=-13 a=6 and c=6
misty1212
  • misty1212
it is not that easy to figure out people claim they have methods, but the methods amount to trial an error anyways
misty1212
  • misty1212
that is why i like to cheat makes life easier this is all donkey work anyways
anonymous
  • anonymous
thank you so much for the help :)

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