Question

if alpha and beta are the zeros of the quadratic polynomial x square - 6 X + a find the valve of a.if 3A+2B = 20​

Answer 1

Appropriate Question :-

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - 6x + a, \: find \: a \: if \: 3 \alpha + 2 \beta = 20$

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

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - 6x + a$

We know,

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

$\bf\implies \: \alpha \beta = \dfrac{a}{1} = a - - - (1)$

and

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

$\bf\implies \: \alpha + \beta = - \dfrac{( - 6)}{1} = 6 - - - (2)$

Also, it is given that

$\rm :\longmapsto\:3 \alpha + 2 \beta = 20 - - - (3)$

Now, from equation (2) we have

$\rm :\longmapsto\: \alpha + \beta = 6$

So, on multiply by 2, we get

$\rm :\longmapsto\: 2\alpha + 2\beta = 12 - - - (4)$

Now, Subtracting equation (4) from equation (3), we get

$\rm :\implies\: \alpha = 8$

On substituting this value in equation (2), we get

$\rm :\longmapsto\:8 + \beta = 6$

$\rm :\longmapsto\:\beta = 6 - 8$

$\rm :\longmapsto\:\beta = - 2$

Hence, we get

$\rm :\longmapsto\: \alpha = 8$

and

$\rm :\longmapsto\:\beta = - 2$

So, substituting these values in equation (1), we get

$\rm :\longmapsto\:a = 8( - 2)$

$\bf\implies \:a \: = \: - \: 16$

Additional Information :-

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

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

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

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