Find the Zeros of h(s) = 2s² − (1 + 2√2)s + √2 and Verify the Relationship Between Zeros and Coefficients

Video Explanation

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Solution

Given polynomial:

h(s) = 2s² − (1 + 2√2)s + √2

Step 1: Find the Zeros of the Polynomial

2s² − (1 + 2√2)s + √2 = 0

Split the middle term:

2s² − s − 2√2s + √2 = 0

Grouping the terms:

s(2s − 1) − √2(2s − 1) = 0

(2s − 1)(s − √2) = 0

∴ 2s − 1 = 0   or   s − √2 = 0

∴ s = 1/2   or   s = √2

Zeros of the polynomial are 1/2 and √2.

Step 2: Identify Coefficients

Comparing h(s) = 2s² − (1 + 2√2)s + √2 with as² + bs + c:

a = 2,   b = −(1 + 2√2),   c = √2

Step 3: Verify the Relationship

Let α = 1/2 and β = √2

Sum of zeros:

α + β = 1/2 + √2

−b/a = −[−(1 + 2√2)] / 2 = (1 + 2√2)/2 = 1/2 + √2

✔ Sum of zeros = −b/a

Product of zeros:

αβ = (1/2)(√2) = √2/2

c/a = √2 / 2 = √2/2

✔ Product of zeros = c/a

Final Answer

Zeros of the polynomial are 1/2 and √2.

The relationship between zeros and coefficients is verified.

Conclusion

Thus, for the quadratic polynomial h(s) = 2s² − (1 + 2√2)s + √2, the sum and product of zeros satisfy the standard relationships with its coefficients.