cdining1 cdining1

15-04-2015
Mathematics

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

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konrad509 konrad509

15-04-2015

[tex]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} [/tex]

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