Question

54. If a and B are the zeroes of 1 point the quadratic polynomial x2 - 6x +k, then the value of k when 3a+2B =20 is ​

Answer 1

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

Given that,

$\rm :\longmapsto\:\alpha, \: \beta \: are \: roots \: of \: quadratic \: polynomial \: {x}^{2} - 6x + k$

and

$\rm :\longmapsto\:3\alpha + 2\beta = 20 - - - (1)$

We know,

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

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

Now, we have from equation (1),

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

$\rm :\longmapsto\:\alpha + 2\alpha + 2\beta = 20$

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

$\rm :\longmapsto\:\alpha + 2 \times 6 = 20$

$\red{\bigg \{ \because \: \alpha + \beta \: =\: 6\bigg \}}$

$\rm :\longmapsto\:\alpha + 12 = 20$

$\rm :\longmapsto\:\alpha = 20 - 12$

$\bf\implies \: \alpha = 8$

Now,

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

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

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

$\bf\implies \:\beta = - 2$

Now, we know that,

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

$\bf\implies \:\alpha\beta = \dfrac{k}{1}$

$\bf\implies \:k = (8)( - 2) = - 16$

Additional Information :-

For cubic polynomial,

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

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

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

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