Question

if alpha and beta are the roots of the polynomial ax^2+bx+c, then find the value of alpha^2+beta^2​

Answer 1

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

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: a {x}^{2} + bx + c}$

We know that,

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

$\bf\implies \: \alpha + \beta = \: - \: \dfrac{b}{a}$

And

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

$\bf\implies \: \alpha \beta = \dfrac{c}{a}$

Now, Consider,

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

can be rewritten as

$\rm \: = \: \: { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta$

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

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

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

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

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

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

Some useful Identities :-

$\boxed{ \rm{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy}}$

$\boxed{ \rm{ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

$\boxed{ \rm{ {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}$