Find the Zeros of p(x) = x² + 2√2x − 6 and Verify the Relationship Between Zeros and Coefficients

Video Explanation

Watch the video explanation below:

Solution

Given polynomial:

p(x) = x² + 2√2x − 6

Step 1: Find the Zeros of the Polynomial

x² + 2√2x − 6 = 0

Split the middle term:

x² + 3√2x − √2x − 6 = 0

Grouping the terms:

x(x + 3√2) − √2(x + 3√2) = 0

(x − √2)(x + 3√2) = 0

∴ x − √2 = 0   or   x + 3√2 = 0

∴ x = √2   or   x = −3√2

Zeros of the polynomial are √2 and −3√2.

Step 2: Identify Coefficients

Comparing p(x) = x² + 2√2x − 6 with ax² + bx + c:

a = 1,   b = 2√2,   c = −6

Step 3: Verify the Relationship

Let α = √2 and β = −3√2

Sum of zeros:

α + β = √2 + (−3√2) = −2√2

−b/a = −(2√2)/1 = −2√2

✔ Sum of zeros = −b/a

Product of zeros:

αβ = (√2)(−3√2) = −6

c/a = −6/1 = −6

✔ Product of zeros = c/a

Final Answer

Zeros of the polynomial are √2 and −3√2.

The relationship between zeros and coefficients is verified.

Conclusion

Thus, for the quadratic polynomial p(x) = x² + 2√2x − 6, the sum and product of zeros satisfy the standard relationships with its coefficients.