Question

$if \alpha and \beta are the zeroes of the polynomial x^{2} -6x+k,find the value of k such that (\alpha +\beta)^{2} -2\alpha \beta =40$

Answer 1

$equation \: is \: {x}^{2} - 6x + k \\ \alpha \: and \: \beta \: are \: rootsof \: this \: eq. \\ \\ so \\ \\ \alpha + \beta = - ( - 6) = 6 \\ \alpha \beta = k \\ \\ directy \: put \: values \: in \: given \: eq. \\ { (\alpha + \beta) }^{2} - 2 \alpha \beta = 40 \\ {(6)}^{2} - 2k = 40 \\ 36 - 2k = 40 \\ - 2k = 4 \\ k = - 2$