Question

if alpha and beta are the zeros of the quadratic polynomial f (x) = x^2 - px +q then evaluate : 1/alpha + 1 /beta​

Answer 1

$\huge{\rm{\underline{\underline{Answer:-}}}}$

The value of $\large{\sf{ \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }}$ is $\large{\sf{ \dfrac{p}{q} }}$.

$\large{\rm{\underline{\bigstar{Step-by-Step\:Explanation:-}}}}$

The given quadratic polynomial is $\large{\sf{f(x) = {x}^{2} - px + q}}$ and its zeroes are $\large{\sf{\alpha}}$ and $\large{\sf{\beta}}$

$\large{\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:{x}{2}}}}$

$\large{\sf{\alpha + \beta = \dfrac{-(-p)}{1}}}$.....(i)

$\large{\sf{ Product\:of\:zeroes = \dfrac{Constant\:term}{Coefficient\:of\:{x}{2}}}}$

$\large{\sf{\alpha \times \beta = \dfrac{q}{1}}}$.....(ii)

Now,

$\large{\sf{ = \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }}$

= $\large{\sf{ \dfrac{\beta + \alpha}{ \alpha \beta } }}$

Using equation (i) and (ii) , put the values of α + β and αβ -

$\large{\sf{ = \dfrac{p}{q} }}$.

Hence the value of $\large{\sf{ \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } }}$ is $\large{\sf{ \dfrac{p}{q} }}$.

Answer 2

Hey there

refer to attachment