Question

find the value of k for kx(x-2)+6=0 has real and equal roots

Answer 1

$hello$


$kx(x - 2) + 6 = 0 \\ \\ = > kx {}^{2} - 2kx + 6 = 0 \\ \\ b {}^{2} - 4ac = 0 \: because \: roots \: are \: real \: and \: equal \\ \\ = > ( - 2kx) {}^{2} - 4 \times kx {}^{2} \times 6 = 0 \\ \\ = > 4kx {}^{2} - 24kx {}^{2} = 0 \\ \\ = > - 20kx {}^{2} = 0 \\ \\ = > k = \frac{0}{ - 20x {}^{2} } \\ \\ = > k = 0$


Therefore,

$\colorbox{orange}{\color{red}{k = 0}}$


HOPE IT HELPS

Answer 2

Hope it helps........