Question

If alpha and beta are zeroes of polynomial f(x)=2x^2+3x+1 then 1/alpha+1/beta =?​

Answer 1

GIVEN :-

• $\rm{f(x) = 2 {x}^{2} + 3x + 1}$
• $\rm{zeroes \: = \alpha \: and \: \beta }$

TO FIND :-

• $\rm{ value \: of \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }$

SOLUTION :-

as we know that for zeroes of a polynomial :-

$\implies \boxed{ \rm{ \alpha + \beta = \dfrac{ - b}{a} }}$

$\implies \boxed{ \rm{ \alpha \times \beta = \dfrac{ c}{a} }}$

where ,

α and β = two zeroes of polynomials

a = coefficient of x²

b = coefficient of x

c = constant term

now applying formula for polynomial 2x²+3x+1

$\implies \rm{ \alpha + \beta = \dfrac{ - 3}{2} \: \: \: \: \: \: \: \: \: \: (1)}$

$\implies \rm{ \alpha \times \beta = \dfrac{ 1}{2} \: \: \: \: \: \: \: \: \: \: \: (2)}$

now we have to find

$\rm{ value \: of \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }$

solve it further :-

$\implies\rm{ \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{ \alpha + \beta }{ \alpha \times \beta } }$

now put the value from eq 1 and eq 2

[tex] \implies \rm{ \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{ \dfrac{ - 3}{2} }{ \dfrac{1}{2} }}[/tex]

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

$\implies \boxed{ \boxed{ \rm{ \dfrac{1}{ \alpha } + \dfrac {1}{ \beta } = - 3 }}}$