Question

Find thr value of k such that the equation (k-12)x square +2(k-12)x+2=0 has equal roots

Answer 1

Answer:

$(k - 12) {x}^{2} + 2(k - 12)x + 2 = 0 \\ equating \: the \: above \:quadratic \: \\ equation \: with \: a {x}^{2} + bx + c = 0 \\ we \: find: \\ a = k - 12 \\ b = 2(k - 12) \\ c = 2 \\ \because \: given \: equation \: has \: equal \: roots \\ \therefore \: {b}^{2} - 4ac = 0 \\ \therefore \: \{2(k - 12) \}^{2} - 4 \times (k - 12) \times 2 = 0 \\ \therefore 4(k - 12)^{2} - 8(k - 12) = 0 \\ \therefore 4(k - 12) \{ (k - 12)-2 \} = 0\\ \therefore (k - 12) \{ k - 12 -2 \} = 0\\ \therefore (k - 12) (k - 14 ) = 0 \\ \therefore k - 12 = 0 \: \: or \: k - 14 = 0 \\ \therefore \: k = 12 \: or \: k = 14$