Question

If alpha and beeta are the zeroes of polynomial 2x^2+7x+5 find
(1/alpha)+(1/beeta)

Answer 1

Answer:

Step-by-step explanation:

2x^2+7x+5

alpha+beeta= -b/a= -7/2

alpha*beeta= c/a= 5/2

(1/alpha)+(1/beeta)

taking LCM,

beeta+alpha/alpha*beeta

that is,

(-7/2)/(5/2)

-7/2*2/5

that is -7/5

Answer 2

Answer:

$\blue{1/alpha+1/beta=-7/5}$

Step-by-step explanation:

$\boxed{given}$

alpha and beta are the zeroes of the polynomial

$\underline{2x^2+7x+5}$

$\boxed{Tofind}$

value of

$\frac{1}{ \alpha } + \frac{1}{ \beta }$

$\boxed{Answer}$

we need to find the sum and the product of the zeroes

which is

$\alpha + \beta = \frac{ - b}{a}$

where

$- b = - (7) = - 7$

$a = 2$

putting the values we get

$\alpha + \beta = \frac{ - 7}{2} ....(1)$

now product of the zeroes

$\alpha \beta = \frac{c}{a}$

where

$c = 5$

$a = 2$

$\alpha \beta = \frac{5}{2} ....(2)$

solving

$\frac{1}{ \alpha } + \frac{1}{ \beta }$

$\frac{ \alpha + \beta }{ \alpha \beta }$

now putting (1) and (2)

we get

$= \frac{ \frac{ - 7}{2} }{ \frac{5}{2} }$

$\boxed{=-7/5}$