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Mathematics 21 Online
OpenStudy (anonymous):

Solve for k: ((16-2k)/3)^2+(2((16-2k)/3))+k=40

OpenStudy (anonymous):

(256-64k+4k^2)/9 + (32-4k)/3 + k =40

OpenStudy (anonymous):

then put them all over 9

OpenStudy (anonymous):

(256-64k+4k^2 + 96-12k+9k-360)/9 = 0

OpenStudy (anonymous):

Wait how'd you get the first step?

OpenStudy (anonymous):

Never mind 16 squared?

OpenStudy (anonymous):

I expanded the first part squared and multiplied the second part by 2

OpenStudy (anonymous):

Do you know what to do from there?

OpenStudy (anonymous):

i think so

OpenStudy (anonymous):

just say so if you need more help

OpenStudy (anonymous):

okay i got to -64k+4k^2-12k+9k=-352, how do i get all the k's alone?

OpenStudy (anonymous):

you can group all the ks together and get 4k^2 + 67k +352 = 0

OpenStudy (anonymous):

then use the quadratic equation with a=4, b=67, c=352

OpenStudy (anonymous):

-67, sorry

OpenStudy (anonymous):

whats the quadratic equation?

OpenStudy (anonymous):

\[k=(-b \pm \sqrt{b ^{2}-4ac})/2a\]

OpenStudy (anonymous):

gotcha!

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