find the zeroes of the quadratic polynomial 7y²-11\3y-2\3 and verify the relationship between the zeroes and their coefficient.
Answers
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 )