Zeroes of the polynomial t? - 11 are *
O+v11
O tv3
O 0
Onone of the above
Answers
Answer:
Step-by-step explanation:
3
2
,
7
−1
Given quadratic polynomial is 7y
2
−
3
11
y−
3
2
.
=
3
1
(21y
2
−11y−2)
=
3
1
(21y
2
−14y+3y−2)
=
3
1
[7y(3y−2)+(3y−2)]
=
3
1
(3y−2)(7y+1)
y=
3
2
, y=−
7
1
The zeroes of the polynomials are,
3
2
, −
7
1
Relationship between the zeroes and the coefficients of the polynomials-
Sum of the zeros=-
coefficient of y
2
coefficient of y
=−
⎝
⎜
⎜
⎜
⎛
7
−
3
11
⎠
⎟
⎟
⎟
⎞
=
21
11
Also sum of zeroes=
3
2
+(−
7
1
)
=
21
14−3
=
21
11
Product of the zeroes =
coefficient of y
2
constant term
=
7
−
3
2
=
21
−2
Also the product of the zeroes=
3
2
×(−
7
1
)=
21
−2
Hence verified.
Option B is correct