Find the zeroes of the quadratic polynomial 7 2 − 11 3 − 2 3 and verify the relationship between the zeroes and their coefficients.
Answers
Answer:
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
Step-by-step explanation:
Given,
q(y) = 7y2 – (11/3)y – 2/3
We put q(y) = 0
⇒ 7y2 – (11/3)y – 2/3 = 0
⇒ (21y2 – 11y -2)/3 = 0
⇒ 21y2 – 11y – 2 = 0
⇒ 21y2 – 14y + 3y – 2 = 0
⇒ 7y(3y – 2) – 1(3y + 2) = 0
⇒ (3y – 2)(7y + 1) = 0
This gives us 2 zeros, for
y = 2/3 and y = -1/7
Hence, the zeros of the quadratic equation are 2/3 and -1/7.
Now, for verification
Sum of zeros = – coefficient of y / coefficient of y2
2/3 + (-1/7) = – (-11/3) / 7
-11/21 = -11/21
Product of roots = constant / coefficient of y2
2/3 x (-1/7) = (-2/3) / 7
– 2/21 = -2/21
Therefore, the relationship between zeros and their coefficients is verified.