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cdining1 cdining1
  • 15-04-2015
  • Mathematics
Answered

21k^2+4=19k solve for k

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konrad509
konrad509 konrad509
  • 15-04-2015

\(21k^2+4=19k\\ 21k^2-19k+4=0\\ 21k^2-7k-12k+4=0\\ 7k(3k-1)-4(3k-1)=0\\ (7k-4)(3k-1)=0\\ k=\dfrac{4}{7} \vee k=\dfrac{1}{3}\)

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