Question

if polynomial is 3x^2-2x-21 then find 1 upon alpha minus 1upon bita

Answer 1

$\boxed{\huge{\bf{\pink{Answer}}}}$

$(\dfrac{1}{\alpha }- \dfrac{1}{\beta })=-\dfrac{16}{21}$

Step-by-step explanation:

$Given \ p(x)=3x^2-2x-21\\\\\\First \ let \ us \ find \ sum \ and \ product \ of \ zeroes\\\\For \ sum\\\\\alpha+\beta=-\dfrac{-b}{a}\\\\\\where \ (b) \ is \ coefficient \ of \ (x) \ and \ (a) \ is \ coefficient \ of \ (x^2)\\\\\alpha+\beta=\dfrac{2}{3}\\\\\\Now \ for \ product\\\\\alpha \beta=\dfrac{c}{a}\\\\\\where \ (c) \ is \ coefficient \ of \ (constant \ term) \ and \ (a) \ is \ coefficient \ of \ (x^2)\\\\\\\alpha \beta=-\dfrac{21}{3}=-7$

$we \ get \ (\alpha+\beta)=\dfrac{2}{3} \ and \ \alpha \beta=-7\\\\\\\(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha \beta\\\\\\(\alpha-\beta)^2=\dfrac{4}{9}+4\times7\\\\\\(\alpha-\beta)^2=\dfrac{4+252}{9}\\\\\\(\alpha-\beta)=\sqrt{\dfrac{256}{9} } =\dfrac{16}{3}$

$we \ have \ to \ find \ (\dfrac{1}{\alpha }- \dfrac{1}{\beta })\\\\\\(\dfrac{1}{\alpha }- \dfrac{1}{\beta })=(\dfrac{\beta -\alpha }{\alpha \beta })\\\\\\(\dfrac{1}{\alpha }- \dfrac{1}{\beta})= \dfrac{-(\alpha-\beta ) }{\alpha \beta }\\\\\\putting \ values \ here\\\\\\(\dfrac{1}{\alpha }- \dfrac{1}{\beta })=\dfrac{-(\frac{16}{3}) }{7}\\\\\\ (\dfrac{1}{\alpha }- \dfrac{1}{\beta })=-\dfrac{16}{21}$