Question

if α and β are the zeros of the polynomial 2x²-7x+3, Find the sum of the
reciprocal of its zeros.

Answer 1

$Given \: \: Question \: \: Is \: \\ p(x) = 2x {}^{2} - 7x + 3 \: \: \: is \: a \: polynomial \: \\ with \: \: \alpha \: \: \: \: and \: \: \: \beta \: \: \: are \: \: its \: \: two \: \: zeros \\ \\ find \: \: \: \: \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ Answer \: \\ \\ p(x) = 2 x{}^{2} - 7x + 3 \\ \\ \alpha + \beta = \frac{7}{2} \: \: \: \: and \: \: \: \: \alpha \beta = \frac{3}{2} \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{( \alpha + \beta )}{ \alpha \beta } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{( \frac{7}{2}) }{ (\frac{3}{2}) } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{3} \\ \\ therefore \: \: sum \: \: of \: \: reciprocal \: \: of \: \: its \: \: zeros \: \: are \\ \\ \frac{7}{3} \\ \\ Note \: \: if \: \: \: p(x) = ax {}^{2} + bx + c\: \: \\ \\ then \: \: \: \alpha + \beta = \frac{ - b}{a} \: \: \: \: and \: \: \: \: \: \alpha \beta = \frac{c}{a} \\ where \: \: \alpha \: \: \: and \: \: \: \beta \: \: are \: \: its \: \: zeros$

Answer 2

Answer:

7 / 3

Step-by-step explanation:

Given :

p ( x ) = 2 x² - 7 x + 3

Sum of zeroes α + β = - b / a

= > -  ( - 7 ) / 2

= > 7 / 2 ... ( i )

Product of zeroes  α β = c / a

= > 3 / 2  ... ( ii )

Now sum of the reciprocal of zeroes :

= > 1 / α + 1 / β

= > ( α + β ) / ( α β )

Putting values from ( i ) and  ( ii ) we get :

= > ( 7 / 2 ) / ( 3 / 2 )

= > 7 / 3

Therefore , sum of the reciprocal of zeroes is 7 / 3 .