Question

if alpha and beta are 2 zeroes of the polynomial. 3x^2-17x+24.find alpha square+beta square and alpha cube+beta cube and 1 by alpha and 1 by beta​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{\alpha\;and\;\beta\;are\;zeros\;of\;3x^2-17x+24}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{1.\alpha^2+\beta^2}$

$\mathsf{2.\alpha^3+\beta^3}$

$\mathsf{3.\dfrac{1}{\alpha}+\dfrac{1}{\beta}}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{3x^2-17x+24}$

$\mathsf{\alpha+\beta=\dfrac{17}{3}}$

$\mathsf{\alpha\beta=\dfrac{24}{3}=8}$

$\mathsf{Now}$

$\mathsf{1.\alpha^2+\beta^2}$

$\mathsf{=(\alpha+\beta)^2-2\alpha\beta}$

$\mathsf{=\left(\dfrac{17}{3}\right)^2-2(8)}$

$\mathsf{=\dfrac{289}{9}-16}$

$\mathsf{=\dfrac{289-144}{9}}$

$\mathsf{=\dfrac{145}{9}}$

$\implies\boxed{\mathsf{\alpha^2+\beta^2=\dfrac{145}{9}}}$

$\mathsf{2.\alpha^3+\beta^3}$

$\mathsf{=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}$

$\mathsf{=\left(\dfrac{17}{3}\right)^3-3(8)\left(\dfrac{17}{3}\right)}$

$\mathsf{=\left(\dfrac{17}{3}\right)^3-8(17)}$

$\mathsf{=\dfrac{4913}{27}-136}$

$\mathsf{=\dfrac{4913-3672}{27}}$

$\mathsf{=\dfrac{1241}{27}}$

$\implies\boxed{\mathsf{\alpha^3+\beta^3=\dfrac{1241}{27}}}$

$\mathsf{3.\dfrac{1}{\alpha}+\dfrac{1}{\beta}}$

$\mathsf{=\dfrac{\alpha+\beta}{\alpha\beta}}$

$\mathsf{=\dfrac{\dfrac{17}{3}}{8}}$

$\mathsf{=\dfrac{17}{24}}$

$\implies\boxed{\mathsf{\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{17}{24}}}$

$\underline{\textsf{Find more:}}$

1)If alpha and beta are zeroes of quadratic polynomial x2 -(k+6x)+2(2k-1) 

find k if alpha +beta=1/2alpha beta  

2)If alpha and beta are zeroes of x2-6x+a, find the value of a if 3alpha+2beta=20

https://brainly.in/question/16419985

3)If α and β are the zeroes of the polynomial

xsquare+ x – 1, the evaluate α + β + αβ

https://brainly.in/question/17511372