Question

Find the value of K, if the roots of the equation (3K + 1)x2 + (11 + K)x + 9 = 0 are equal

Answer 1

$\text{since, \:roots\: are \:equal.}$

$hence, D = 0$

$so,$

$here a = (3k+1), b = (11+k), c = 9$

$then,$

$D = 0$

${b}^{2}-4ac = 0$

${(11+k)}^{2}-4×(3k+1)×9 = 0$

${(11)}^{2}+{k}^{2}+2×11×k-4×9(3k+1) = 0$

$1331 + {k}^{2}+ 22k -36(3k+1) = 0$

$1331 + {k}^{2}+ 22k -108k -36 = 0$

${k}^{2} +22k-108k+1331-36 = 0$

${k}^{2} -86k + 1295 = 0$

$\text{Now, \:mid \:term \:split}$

$\text{we \:get,}$

Your question is little bit wrong I think so.