Question

For equal root, kx(x-2) +6=0, value of k is

Answer 1

Step-by-step explanation:

$kx(x - 2) + 6 = 0 \\ k {x}^{2} - 2kx + 6 = 0 \\ a = k \: \: b = - 2k \: \: c = 6\\ given \\ there \: are \: equal \: roots \\ = > {b}^{2} - 4ac = 0 \\ = > { (- 2k)}^{2} - 4(k)(6) = 0 \\ = > 4 {k}^{2} - 24k = 0 \\ = >4k(k - 6) = 0 \\ = > 4k = 0 \: \: \: \: \: k - 6 = 0 \\ = > k = \frac{0}{4} \: \: \: \: \: \: k = 6 \\ = > k = 0 \: \: \: \: \: \: \: k = 6 \\ therefore \\ k = 0 \: or \: 6$