If[tex] \alpha [/tex]
and
[tex] \beta [/tex]
are the zeroes of tge zeroes of the polynomial f (x) = x² - 5x +k such that
[tex] \alpha - \beta = 1[/tex]
find the value of k
Share
If[tex] \alpha [/tex]
and
[tex] \beta [/tex]
are the zeroes of tge zeroes of the polynomial f (x) = x² - 5x +k such that
[tex] \alpha - \beta = 1[/tex]
find the value of k
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\rm :\longmapsto\: \alpha , \beta \: are \: the \: zeroes \: of \: {x}^{2} - 5x + k[/tex]
We know,
[tex]\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\rm :\implies\: \alpha + \beta = - \: \dfrac{( - 5)}{1} = 5[/tex]
Also,
[tex]\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}[/tex]
[tex]\rm :\implies\: \alpha \beta = \dfrac{k}{1} = k[/tex]
Also, it is given that
[tex]\rm :\longmapsto\: \alpha - \beta = 1[/tex]
On squaring both sides, we get
[tex]\rm :\longmapsto\: (\alpha - \beta)^{2} = 1[/tex]
We know,
[tex]\underbrace{\boxed{ \tt{ {(x + y)}^{2} - {(x - y)}^{2} \: = \: 4xy}}}[/tex]
So, using this property, we get
[tex]\rm :\longmapsto\: (\alpha + \beta)^{2} - 4 \alpha \beta = 1[/tex]
On substituting the values, we get
[tex]\rm :\longmapsto\: {5}^{2} - 4k = 1[/tex]
[tex]\rm :\longmapsto\: 25 - 4k = 1[/tex]
[tex]\rm :\longmapsto\: - 4k = 1 - 25[/tex]
[tex]\rm :\longmapsto\: - 4k = - 24[/tex]
[tex]\bf\implies \:k = 6[/tex]
Additional Information :-
[tex]\rm :\longmapsto\: \alpha , \beta, \gamma \: are \: the \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d, \: then[/tex]
[tex]\underbrace{\boxed{ \tt{ \alpha + \beta + \gamma = - \: \frac{b}{a} \: }}}[/tex]
[tex]\underbrace{\boxed{ \tt{ \alpha \beta + \beta \gamma + \gamma \alpha = \: \frac{c}{a} \: }}}[/tex]
[tex]\underbrace{\boxed{ \tt{ \alpha \beta \gamma = - \: \frac{d}{a} \: }}}[/tex]