Find the Zeros of the Following Quadratic Polynomials x²+7x+10 and Verify the Relationship between the Zeros and the Coefficients

Find the Zeros of the Following Quadratic Polynomials x²+7x+10 and Verify the Relationship between the Zeros and the Coefficients

Solution:

To find the zeros of the quadratic polynomial x² + 7x + 10, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Comparing the polynomial to the standard quadratic form ax² + bx + c, we have a = 1, b = 7, and c = 10. Substituting these values into the quadratic formula, we get:

x = (-(7) ± √((7)² - 4(1)(10))) / (2(1))

Simplifying further:

x = (-7 ± √(49 - 40)) / 2

x = (-7 ± √9) / 2

x = (-7 ± 3) / 2 = -2

This gives us two possible solutions:

x₁ = (-7 + 3) / 2 = -2

x₂ = (-7 - 3) / 2 = -5

Therefore, the zeros of the quadratic polynomial x² + 7x + 10 are x = -2 and x = -5.

Now, let's verify the relationship between the zeros and the coefficients using Vieta's formulas:

The sum of the zeros is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term:

Sum of zeros = -(7/1) / (1/1) = -7

The product of the zeros is equal to the constant term divided by the coefficient of the quadratic term:

Product of zeros = (10/1) / (1/1) = 10

Indeed, the sum of the zeros, -7, matches the coefficient of the linear term, and the product of the zeros, 10, matches the constant term. Therefore, the relationship between the zeros and the coefficients is verified.

Answer:

Zeros of the Following Quadratic Polynomials x²+7x+10 are -2 and -5 which are Verified Correctly

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