Find the Zeros of p(y) = y² + (3√5/2)y − 5 and Verify the Relationship Between Zeros and Coefficients

Video Explanation

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Solution

Given polynomial:

p(y) = y² + (3√5/2)y − 5

Step 1: Find the Zeros of the Polynomial

y² + (3√5/2)y − 5 = 0

Multiply the whole equation by 2 to remove the fraction:

2y² + 3√5y − 10 = 0

Split the middle term:

2y² + 4√5y − √5y − 10 = 0

Grouping the terms:

2y(y + 2√5) − √5(y + 2√5) = 0

(2y − √5)(y + 2√5) = 0

∴ 2y − √5 = 0   or   y + 2√5 = 0

∴ y = √5/2   or   y = −2√5

Zeros of the polynomial are √5/2 and −2√5.

Step 2: Identify Coefficients

Comparing p(y) = y² + (3√5/2)y − 5 with ay² + by + c:

a = 1,   b = 3√5/2,   c = −5

Step 3: Verify the Relationship

Let α = √5/2 and β = −2√5

Sum of zeros:

α + β = √5/2 − 2√5 = −3√5/2

−b/a = −(3√5/2)/1 = −3√5/2

✔ Sum of zeros = −b/a

Product of zeros:

αβ = (√5/2)(−2√5) = −5

c/a = −5/1 = −5

✔ Product of zeros = c/a

Final Answer

Zeros of the polynomial are √5/2 and −2√5.

The relationship between zeros and coefficients is verified.

Conclusion

Thus, for the quadratic polynomial p(y) = y² + (3√5/2)y − 5, the sum and product of zeros satisfy the standard relationships with its coefficients.