Question

if α and β are roots of the equation x^2+6x+c=0 and 3α+2β=-20 then c=

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} + 6x + c = 0$

and

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

Now, it is given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {x}^{2} + 6x + c = 0$

We know,

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

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

As it is given that,

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

can be rewritten as

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

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

$\rm :\longmapsto\:2( - 6 ) + \alpha = - 20$

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

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

$\rm \implies\:\boxed{ \tt{ \: \alpha \: = \: - \: 8 \: }} - - - (2)$

On substituting equation (2) in equation (1), we get

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

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

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

$\rm :\longmapsto\: \boxed{ \tt{ \: \beta = 2 \: }} - - - (3)$

Also, we know that,

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

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

$\bf\implies \:(2)( - 8) = c$

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

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Additional Information :-

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

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

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

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