Question

If α and β are the zeroes of the polynomial ax2 + bx + c, find the value of α2 + β2. (2013)
Solution​

Answer 1

Given :-

• α and β are the zeroes of the polynomial ax² + bx + c

To find :-

• The value of α² + β²

Solution :-

Given that,

• α and β are the zeroes of the polynomial ax² + 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,

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

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

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

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

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

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{\boxed{\bf \: { \alpha }^{2} + { \beta }^{2} = \dfrac{ {b}^{2} - 2ac }{ {a}^{2}}}}$

Additional Information :-

Important Identities :-

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

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

Cubic Polynomial :-

$\sf \: If \: \alpha , \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d \: then$

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

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

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