Question

Find the value of k, the equation has equal roots x ²-K₂+(K-1) = 0​

Answer 1

Answer:

the value of k =2

Step-by-step explanation:

${x}^{2} - kx + (k - 1) = 0 \\ discriminat \: formula \\ a {x}^{2} + bx + c = 0 \\ then \\ d = {b}^{2} - 4ac \\ so \: \: \: \: \: a = 1 \: \: b = -k \: \: \: c = (k - 1) \\ d = ( { - k})^{2} - 4(1)(k - 1) \\ = \: {k}^{2} - 4k + 4 \\ if \: there \: are \: equal \: roots \: then \: value \: of \: d = 0 \\ {k}^{2} - 4k + 4 = 0 \\ {k}^{2} - 2k - 2k + 4 = 0 \\ k(k - 2) - 2(k - 2) = 0 \\ (k - 2)(k - 2) = 0 \\ {(k - 2)}^{2} = 0 \\ (k - 2) = 0 \\ k - 2 = 0 \\ k = 2$