Question

if alpha and beta are the zeroes of the polynomial 3x^2 + 5x = 2 then find, the value of 1/alpha +1/beta​

Answer 1

Given :

$\alpha \: and \: \beta \: are \: the \: zeroes \\ \: of \: the \: polynomial \: 3 {x}^{2} + 5x = 2$

To find :

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

Solution :

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

$LCM \: = {\alpha} {\beta}$

$\implies\dfrac{\beta + \alpha}{\alpha\beta}$

$\implies\dfrac{\beta + \alpha}{\alpha\beta} = \dfrac{sum \: of \: roots}{product \: of \: root}$

$\implies \: 3 {x}^{2} + 5x = 2$

$\implies \: 3 {x}^{2} + 5x - 2 = 0$

$sum \: of \: roots = - \: \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\implies \: sum \: of \: roots = - \: \dfrac{5}{3} = {\beta + \alpha}$

$product\: of \: roots = - \: \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\implies \: product \: of \: roots = \: \dfrac{ - 2}{3 } = {\alpha\beta}$

Now :-

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

$\implies\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{ - \: \dfrac{5}{3} }{ \dfrac{ - 2}{3 } }$

$\implies\dfrac{1}{\alpha} + \dfrac{1}{\beta} = (\dfrac{ - 5}{3}) \times ( \dfrac{ - 3}{2} )$

$\implies\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{ 5}{2}$

Answer :

$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{ 5}{2}$