Question

if alpha and beta are the zeros of the polynomial x2-3x+6 then find the value of 1/alpha+1/beta+alpha2 + beta2 - 2alphabeta

Answer 1

Answer is( - 29/2)

STEP-BY-STEP EXPLANATION :-

$p(x) = {x}^{2} - 3x + 6 \\ \\ \\ \alpha \: and \: \beta \: are \: zeroes \\ \\ \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \\ \alpha \beta = \frac{c}{a}$

Now,

$\frac{1}{ \alpha } + \frac{1}{ \beta } + { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta \\ \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } + { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta + 4 \alpha \beta - 4 \alpha \beta \\ \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } + { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 4 \alpha \beta \\ \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } + {( \alpha + \beta )}^{2} - 4 \alpha \beta$

Now,

Putting down the values :-

$\dfrac{ \dfrac{ - b}{a}} { \dfrac{c}{a} } + {( \dfrac{ - b}{a}) }^{2} - 4( \dfrac{c}{a} ) \\ \\ \\ = \frac{ - b}{c} + {( \frac{ - b}{a}) }^{2} - 4( \frac{c}{a} ) \\ \\ \\ = \frac{3}{6} + {(3)}^{2} - 4 \times 6 \\ \\ = \frac{3}{6} + 9 - 24 \\ \\ = \frac{1}{2} - 15\\ \\ =\frac{1-30}{2}\\ =\frac{-29}{2}$