Question

10) If alpha and beta are the zeroes of the polynomial f(x) = x2 - 4x + k such that alpha-beta = 2, find the value of k​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {x}^{2} - 4x + k$

We know,

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

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

And

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

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

Now, According to statement

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

Squaring both sides, we get

$\rm :\longmapsto\: {( \alpha - \beta) }^{2} = 4$

$\rm :\longmapsto\: {( \alpha + \beta) }^{2} - 4 \alpha \beta = 4$

$\red{\bigg \{ \because \: {( x- y)}^{2} = {(x + y)}^{2} - 4xy\bigg \}}$

$\rm :\longmapsto\: {4}^{2} - 4k = 4$

$\rm :\longmapsto\: 16 - 4k = 4$

$\rm :\longmapsto\: - 4k = 4 - 16$

$\rm :\longmapsto\: - 4k = - 12$

$\bf\implies \:k = 3$

ADDITIONAL INFORMATION :-

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

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

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

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