Question

if alpha and beta are the roots of equation (-2x²+6x-3) then alpha square+ beta square equals?​

Answer 1

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

$\rm :\longmapsto\:\sf \: Since \: \alpha \: and \: \beta \: are \: roots \: of \: - {2x}^{2} + 6x - 3$

We know that,

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

$\bf\implies \: \alpha + \beta = - \dfrac{6}{( - 2)} = 3$

And

$\: \: \: \: \: \boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

$\bf\implies \: \alpha \beta = \dfrac{ - 3}{ - 2} = \dfrac{3}{2}$

Consider,

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

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

$\sf \: = \: \: \: {(3)}^{2} - 2 \times \dfrac{3}{2}$

$\sf \: = \: \: \:9 - 3$

$\sf \: = \: \: \:6$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf\: { \alpha }^{2} + { \beta }^{2} = 6}}$

Additional Information :-

$\rm :\longmapsto\:\sf \: If \: \alpha, \: \beta \: and \: \gamma \: are \: roots \: of \: {ax}^{3} + {bx}^{2} + cx + d$

then,

$\rm :\longmapsto\: \alpha + \beta + \gamma = - \dfrac{b}{a}$

$\rm :\longmapsto\: \alpha \: \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}$

$\rm :\longmapsto\: \alpha\beta \gamma = - \dfrac{d}{a}$