Question

if alpha and beta are the zeros of the polynomial f of x = 5 x square + 4 x minus 9 then evaluate Alpha square + beta square​

Answer 1

Step-by-step explanation:

Given polynomial is:

$p(x) = 5 {x}^{2} + 4x - 9 \\ equting \: it \: with \\ p(x) = a {x}^{2} + bx + c \\ we \: find \: \\ a \: = 5 \\ b = 4 \\ c = - 9 \\ \because \: \alpha \: and \: \beta \: are \: zeros \: of \: given \\ \: polynomial \\ \therefore \: sum \: of \: zeros \\ \alpha + \beta = - \frac{b}{a} = - \frac{4}{5} \\ product \: of \: zeros \\ \alpha \beta = \frac{c}{a} = \frac{ - 9}{5} \\ now \\ ( { \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta \\ ( - \frac{4}{5} ) ^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \times ( \frac{ - 9}{5} ) \\ \frac{16}{25} = { \alpha }^{2} + { \beta }^{2} - \frac{18}{5} \\ \frac{16}{25} + \frac{18}{5} = { \alpha }^{2} + { \beta }^{2} \\ \frac{16}{25} + \frac{18 \times 5}{5 \times 5} = { \alpha }^{2} + { \beta }^{2} \\ { \alpha }^{2} + { \beta }^{2} = \frac{16}{25} + \frac{90}{25} = \frac{16 + 90}{25} \\ { \alpha }^{2} + { \beta }^{2} = \frac{106}{25}$