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Find the value of k so that the quadratic equation ky(y-7)+49=0 has two equal roots

Answer» HEYA!!
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KY ( y-7 ) + 49

ky {}^{2} - 7ky + 49 \\ \\ for \: the \: equation \: to \: have \: equal \: roots \: \\ \\ b {}^{2} - 4ac = 0 \\ \\ ( - 7k) {}^{2} - <klux>4</klux> \times <klux>K</klux> \times 49 \\ \\ 49k {}^{2} - 196k = 0 \\ \\ 7k(7k - 28) = 0 \\ \\ 7k - 28 = 0 \\ \\ 7(k - 4) = 0 \\ \\ k = > 4
Hence the VALUE of K = 4

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