Question

if alpha and beta are the zeros of the polynomal f(x)=x2-6x+k,find the values of k such that alpha2+beta2=40

Answer 1

$since \: \alpha \: and \: \beta \: are \: the \: roots \\ { \alpha }^{2} - 6 \alpha + k = 0 - - - - - (1) \\ { \beta }^{2} - 6 \beta + k = 0 - - - - - (2)$
Adding (1) and (2),
$( { \alpha }^{2} + { \beta }^{2} ) - 6( \alpha + \beta ) + 2k = 0 - - - (3)$
$from \: the \: quadratic \: equation \\ sum \: of \: the \: roots \\ \alpha + \beta = 6 - - - - (4)$
$it \: is \: given \: { \alpha }^{2} + { \beta }^{2} = 40 - - (5)$
$substituting \: (4) \: and \: (5) \: in \: (3) \\ 40 - 6 \times 6 + 2k = 0 \\ k = - 2$