Question

If Alfa , beta are the roots of x^2-x+2 = 0, then alfa^2Beta + alfa beta^2 is​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} - x + 2 = 0$

We know,

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

$\rm \implies\: \alpha + \beta = - \dfrac{( - 1)}{1} = 1$

and

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

$\rm \implies\: \alpha \beta = \dfrac{2}{1} = 2$

Now,

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

$\rm \:  =  \: \alpha \beta ( \alpha + \beta )$

So, on substituting the values, we get

$\rm \:  =  \:1 \times 2$

$\rm \:  =  \:2$

Hence,

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

More to know :-

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

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

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

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

Answer 2

Step-by-step explanation:

alfa^2Beta + alfa beta^2 = 2

Hope it helps you