anonymous
  • anonymous
Solve for k: ((16-2k)/3)^2+(2((16-2k)/3))+k=40
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
(256-64k+4k^2)/9 + (32-4k)/3 + k =40
anonymous
  • anonymous
then put them all over 9
anonymous
  • anonymous
(256-64k+4k^2 + 96-12k+9k-360)/9 = 0

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anonymous
  • anonymous
Wait how'd you get the first step?
anonymous
  • anonymous
Never mind 16 squared?
anonymous
  • anonymous
I expanded the first part squared and multiplied the second part by 2
anonymous
  • anonymous
Do you know what to do from there?
anonymous
  • anonymous
i think so
anonymous
  • anonymous
just say so if you need more help
anonymous
  • anonymous
okay i got to -64k+4k^2-12k+9k=-352, how do i get all the k's alone?
anonymous
  • anonymous
you can group all the ks together and get 4k^2 + 67k +352 = 0
anonymous
  • anonymous
then use the quadratic equation with a=4, b=67, c=352
anonymous
  • anonymous
-67, sorry
anonymous
  • anonymous
whats the quadratic equation?
anonymous
  • anonymous
\[k=(-b \pm \sqrt{b ^{2}-4ac})/2a\]
anonymous
  • anonymous
gotcha!

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