Find the Zeros of q(y) = 7y² − (11/3)y − 2/3 and Verify the Relationship Between Zeros and Coefficients

Video Explanation

Watch the video explanation below:

Solution

Given polynomial:

q(y) = 7y² − (11/3)y − 2/3

Step 1: Find the Zeros of the Polynomial

7y² − (11/3)y − 2/3 = 0

Multiply the whole equation by 3 to remove fractions:

21y² − 11y − 2 = 0

Split the middle term:

21y² − 14y + 3y − 2 = 0

Grouping the terms:

7y(3y − 2) + 1(3y − 2) = 0

(7y + 1)(3y − 2) = 0

∴ 7y + 1 = 0   or   3y − 2 = 0

∴ y = −1/7   or   y = 2/3

Zeros of the polynomial are −1/7 and 2/3.

Step 2: Identify Coefficients

Comparing q(y) = 7y² − (11/3)y − 2/3 with ay² + by + c:

a = 7,   b = −11/3,   c = −2/3

Step 3: Verify the Relationship

Let α = −1/7 and β = 2/3

Sum of zeros:

α + β = −1/7 + 2/3

= (−3 + 14)/21 = 11/21

−b/a = −[−11/3] / 7 = 11/21

✔ Sum of zeros = −b/a

Product of zeros:

αβ = (−1/7)(2/3) = −2/21

c/a = (−2/3) / 7 = −2/21

✔ Product of zeros = c/a

Final Answer

Zeros of the polynomial are −1/7 and 2/3.

The relationship between zeros and coefficients is verified.

Conclusion

Thus, for the quadratic polynomial q(y) = 7y² − (11/3)y − 2/3, the sum and product of zeros satisfy the standard relationships with its coefficients.