Question

if alpha and beta are the zeros of quratic polynomial p(x)= 6x^2+x-2 then find the value of 1/alpha+1/beta​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\bf Given:\ \sf \alpha\ and\ \beta\ are\ the\ zeroes\ of\ the\ polynomial\ 6x^2\ +\ x\ -\ 2.\\\\\bf To\ find:\ \sf The\ value\ of\ \dfrac{1}{\alpha}\ +\ \dfrac{1}{\beta}.\\\\\bf Answer:\\\\\sf From,\ the\ equation:\ a\ =\ 6,\ b\ =\ 1,\ c\ =\ -2.\\\\We\ know\ that\ the\ sum\ of\ the\ zeroes\ of\ a\ quadratic\ polynomial\ =\ \dfrac{-b}{a}.\\\\\\\implies\ \alpha\ +\ \beta\ =\ \dfrac{-1}{6}\\\\\\We\ also\ know\ that\ the\ product\ of\ the\ zeroes\ is\ \dfrac{c}{a}.$

$\sf \implies\ \alpha\ \times\ \beta\ =\ \dfrac{-2}{6}\ =\ \dfrac{-1}{3}\\\\\\\\\dfrac{1}{\alpha}\ +\ \dfrac{1}{\beta}\ \\\\\\=\ \dfrac{\alpha\ +\ \beta}{\alpha\ \times\ \beta}\\\\\\=\ \dfrac{-1}{6}\ \times\ -3\\\\\\=\ \dfrac{3}{6}\\\\\\=\ \dfrac{1}{2}$