Question

if alpha and beta are the zeroes of quadratic polynomial ax sqauare + bx + c evaluate alpha by beta + beta by alpha​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \bull \: \sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:{{\tt Sum\ of\ the\ zeroes \: ( \alpha + \beta ) \: = \: \dfrac{-b}{a}}}$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:{{\tt Product\ of\ the\ zeroes \: ( \alpha \beta )=\dfrac{c}{a}}}$

Now, Consider

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2}$

$\sf \: = \: {( \alpha + \beta )}^{2} - 2 \alpha \beta$

$\sf \: = \: {\bigg(\dfrac{ - b}{a} \bigg) }^{2} - 2\bigg(\dfrac{c}{a} \bigg)$

$\sf \: = \: \dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{2c}{a}$

$\sf \: = \: \dfrac{ {b}^{2} - 2ac }{ {a}^{2} }$

Onwards this, please find the attachment

Additional Information :-

$\boxed{ \bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }$

$\boxed{ \bf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta )}$

$\boxed{ \bf \: \alpha - \beta = \sqrt{ {( \alpha + \beta )}^{2} - 4 \alpha \beta } }$