Question

If$\alpha$
and
$\beta$
are the zeroes of tge zeroes of the polynomial f (x) = x² - 5x +k such that
$\alpha - \beta = 1$
find the value of k ​

Answer 1

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

Given that

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: zeroes \: of \: {x}^{2} - 5x + k$

We know,

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

$\rm :\implies\: \alpha + \beta = - \: \dfrac{( - 5)}{1} = 5$

Also,

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

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

Also, it is given that

$\rm :\longmapsto\: \alpha - \beta = 1$

On squaring both sides, we get

$\rm :\longmapsto\: (\alpha - \beta)^{2} = 1$

We know,

$\underbrace{\boxed{ \tt{ {(x + y)}^{2} - {(x - y)}^{2} \: = \: 4xy}}}$

So, using this property, we get

$\rm :\longmapsto\: (\alpha + \beta)^{2} - 4 \alpha \beta = 1$

On substituting the values, we get

$\rm :\longmapsto\: {5}^{2} - 4k = 1$

$\rm :\longmapsto\: 25 - 4k = 1$

$\rm :\longmapsto\: - 4k = 1 - 25$

$\rm :\longmapsto\: - 4k = - 24$

$\bf\implies \:k = 6$

Additional Information :-

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

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

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

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