Question

Please answer it questio no 14

Answer 1

$given \: polynomial\: is \\ \\ f(x) = x {}^{2} - 6x + k \\ \\ therefore \: sum \: of \: the \: roots \: is \\ \\ \alpha + \beta = - \frac{ - 6}{1} = 6 \\ \\ and \: product \: of \: roots \: is \\ \\ \alpha \beta = \frac{k}{1} = k \\ \\ since \\ \\ ( \alpha + \beta ) {}^{2} = \alpha {}^{2} + \beta {}^{2} +2 \alpha \beta .........(2) \\ \\ putting \: the \: valuesof \: ( \alpha + \beta ) \: and \: \alpha \beta in \: equation \: (2) \\ \\ 6 {}^{2} = \alpha {}^{2} + \beta {}^{2} + 2 k \\ \\ 36 = \alpha {}^{2} + \beta {}^{2} + 2k \\ \\ but \: given \: that \: \alpha {}^{2} + \beta {}^{2} = 40 \\ \\ therefore \\ \\ 36 = 40 + 2k \\ \\ 2k = 40 - 36 \\ \\ 2k = - 4 \\ \\ k = \frac{ - 4}{2} \\ \\ k = - 2 \\ \\ hope \: it \: helps...$