Question

find the value of k for each of the following quadratic equation so that they have two equal roots x²-6x+(k-2)=0​

Answer 1

Step-by-step explanation:

for equal roots

${b}^{2} - 4ac = 0 \\ for \: a \: quadratic \: equation \: ax^{2} + bx + c$=0

$given \: equation \: {x}^{2} - 6x + (k - 2) \\ comparing \: by \: a {x}^{2} + bx + c = 0 \\ a = 1 \\ b = - 6 \\ c = k - 2 \\ using \: {b}^{2} - 4c = 0 \\ ( - 6)^{2} - 4 \times 1 \times (k - 2) = 0 \\ 36 - 4k + 8 = 0 \\ 44 = 4k \\ \frac{44}{4 } = k \\ k = 11$