Question

If alpha and beta are the zeroes of the x²-x-4, then find out the value of one upon alpha plus one upon beta minus alpha multiply by beta​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - x - 4$

We know that,

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

$\rm :\implies\: \alpha \beta = \dfrac{ - 4}{1} = - 4$

and

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

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

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } - \alpha \beta$

$\rm \: = \:\dfrac{ \beta + \alpha }{ \alpha \beta } - ( - 4)$

$\rm \: = \:\dfrac{1}{ - 4} + 4$

$\rm \: = - \:\dfrac{1}{ 4} + 4$

$\rm \: = \:\dfrac{ - 1 + 16}{ 4}$

$\rm \: = \:\dfrac{15}{ 4}$

Hence,

$\rm :\longmapsto\: \boxed{ \bf{ \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } - \alpha \beta = \frac{15}{4}}}$

Additional Information :-

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

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

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

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

Answer 2

Answer:

Ahh... miss did we talk before... ah I forgot -_-||

Sorry for that... are u a K-pop fan or an anime fan

(￣3￣)