Hi ,

Let p( y ) = 7y² - 11y/3 - 2/3 ,

To find the zeroes , we have to take

p ( y ) = 0

7y² - 11y/3 - 2/3 = 0

Multiply each term with ' 3 ' we get

21y² - 11y - 2 = 0

21y² - 14y + 3y - 2 = 0

7y ( 3y - 2 ) + 1( 3y - 2 ) = 0

( 3y - 2 ) ( 7y + 1 ) = 0

Therefore ,

3y - 2 = 0 or 7y + 1 = 0

3y = 2 or 7y = -1

y = 2/3 or y = ( -1/7 )

Therefore ,

Required two zeroes of p( y ) are

m = 2/3 , n = -1/7

************************

Compare p( y ) with ax² + bx + c , we

get

a = 7 , b = -11/3 , c = 2/3 ,

1 ) sum of the zeroes = -b / a

= - ( -11/3 )/ 7

= 11/21 ----( 1 )

m + n = 2/3 - 1/7

= ( 14 - 3 ) / 21

= 11/21 ---( 2 )

( 1 ) = ( 2 )

2 ) product of the zeroes = c/a

= ( -2/3 ) /7

= - 2/21----( 3 )

mn = (2/3 ) ( - 1/7 )

=- 2/21 ----( 4 )

(3 ) = (4 )

I hope this helps you.

:)

Answer:

it is easy

Step-by-step explanation:

Let p( y ) = 7y² - 11y/3 - 2/3 ,

To find the zeroes , we have to take

p ( y ) = 0

7y² - 11y/3 - 2/3 = 0

Multiply each term with ' 3 ' we get

21y² - 11y - 2 = 0

21y² - 14y + 3y - 2 = 0

7y ( 3y - 2 ) + 1( 3y - 2 ) = 0

( 3y - 2 ) ( 7y + 1 ) = 0

Therefore ,

3y - 2 = 0 or 7y + 1 = 0

3y = 2 or 7y = -1

y = 2/3 or y = ( -1/7 )

Therefore ,

Required two zeroes of p( y ) are

m = 2/3 , n = -1/7

Compare p( y ) with ax² + bx + c , we

get

a = 7 , b = -11/3 , c = 2/3 ,

1 ) sum of the zeroes = -b / a

= - ( -11/3 )/ 7

= 11/21 ----( 1 )

m + n = 2/3 - 1/7

= ( 14 - 3 ) / 21

= 11/21 ---( 2 )

( 1 ) = ( 2 )

2 ) product of the zeroes = c/a

= ( -2/3 ) /7

= - 2/21----( 3 )

mn = (2/3 ) ( - 1/7 )

=- 2/21 ----( 4 )

(3 ) = (4 )