Question

Answer fast.i will mark u brainliest​

Answer 1

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

Given that

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: k {x}^{2} + 4x + 4$

We know,

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

$\rm :\implies\: \alpha + \beta = - \: \dfrac{4}{k} - - - (1)$

and

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

$\rm :\implies\: \alpha \beta \: =\: \dfrac{4}{k} - - - (2)$

Now, further it is given that

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2} = 24$

$\rm :\longmapsto\: {(\alpha + \beta ) }^{2} - 2 \alpha \beta = 24$

$\rm :\longmapsto\: {\bigg( - \dfrac{4}{k} \bigg) }^{2} - 2 \times \dfrac{4}{k} = 24$

$\rm :\longmapsto\:\dfrac{16}{ {k}^{2} } - \dfrac{8}{k} = 24$

$\rm :\longmapsto\:\dfrac{16 - 8k}{ {k}^{2} } = 24$

On dividing by 8, we get

$\rm :\longmapsto\:\dfrac{2 - k}{ {k}^{2} } = 3$

$\rm :\longmapsto\:2 - k = {3k}^{2}$

$\rm :\longmapsto\: {3k}^{2} + k - 2 = 0$

$\rm :\longmapsto\: {3k}^{2} + 3k - 2k - 2 = 0$

$\rm :\longmapsto\:3k(k + 1) - 2(k + 1) = 0$

$\rm :\longmapsto\:(k + 1)(3k - 2) = 0$

$\rm :\implies\:k = - 1 \: \: \: or \: \: \: k = \dfrac{2}{3}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (B) \: is \: correct}}}$

Additional Information :-

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

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \: \frac{b}{a}}}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \frac{c}{a}}}$

$\boxed{ \sf{ \: \alpha\beta \gamma = - \: \frac{d}{a}}}$