Question

if alpha and beta are the zeros of the polynomial 3x^2 - 4x - 7 then form a quadratic equation whose zeroes are 1/alpha and 1/beta

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\alpha\;and\;\beta\;are\;zeros\;of\;the\;polynomial\;3x^2-4x-7}$

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

$\textsf{A quadratic equation whose zeroes are}$

$\mathsf{\dfrac{1}{\alpha}\;and\;\dfrac{1}{\beta}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;3x^2-4x-7}$

$\mathsf{Sum\;of\;zeroes=\dfrac{-b}{a}=\dfrac{-(-4)}{3}=\dfrac{4}{3}}$

$\implies\mathsf{\alpha+\beta=\dfrac{4}{3}}$

$\mathsf{Product\;of\;zeroes=\dfrac{c}{a}=\dfrac{-7}{3}=\dfrac{-7}{3}}$

$\implies\mathsf{\alpha\,\beta=\dfrac{-7}{3}}$

$\mathsf{Now,}$

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

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

$\mathsf{=\dfrac{\dfrac{4}{3}}{\dfrac{-7}{3}}}$

$\mathsf{=\dfrac{-4}{7}}$

$\mathsf{\dfrac{1}{\alpha}{\times}\dfrac{1}{\beta}}$

$\mathsf{=\dfrac{1}{\alpha\beta}}$

$\mathsf{=\dfrac{1}{\dfrac{-7}{3}}}$

$\mathsf{=\dfrac{-3}{7}}$

$\underline{\textsf{The required polynomial is}}$

$\mathsf{=x^2-\left(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right)x+\left(\dfrac{1}{\alpha}\,\dfrac{1}{\beta}\right)}$

$\mathsf{=x^2-\left(\dfrac{-4}{7}\right)x+\left(\dfrac{-3}{7}\right)}$

$\mathsf{=x^2+\dfrac{4}{7}x-\dfrac{3}{7}}$

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

if α & β are the zeros of the quadratic polynomial p(x)= 3X²-6x+4,find the value of α\β+β+α+2(1/α+1/β)+3αβ​

https://brainly.in/question/18209208