Question

Which of the following quadratic polynomials has zeroes \( \frac{3}{5} \) and \( -\frac{1}{2} \)?

• A. \( 10x^2 + x + 3 \)
• B. \( 10x^2 - x - 3 \)
• C. \( 10x^2 - x + 3 \)
• D. \( 10x^2 - x - 3 \)

Hint:

Given the zeroes of a quadratic polynomial, you can reconstruct the polynomial using the sum and product of the zeroes.

Answer 1

The Correct Option is B

To find the quadratic polynomial from its zeroes \( \alpha = \frac{3}{5} \) and \( \beta = -\frac{1}{2} \), use the fact that the sum and product of the zeroes of a quadratic polynomial \( ax^2 + bx + c \) are given by: \[ \text{Sum of zeroes} = -\frac{b}{a}, \quad \text{Product of zeroes} = \frac{c}{a}. \] The sum of the zeroes is: \[ \alpha + \beta = \frac{3}{5} + \left( -\frac{1}{2} \right) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}. \] The product of the zeroes is: \[ \alpha \cdot \beta = \frac{3}{5} \cdot \left( -\frac{1}{2} \right) = -\frac{3}{10}. \] Now, we can write the quadratic polynomial as: \[ a(x^2 - (\alpha + \beta)x + \alpha \beta). \] Substituting the sum and product of the zeroes: \[ a(x^2 - \frac{1}{10}x - \frac{3}{10}). \] Multiplying through by 10 to eliminate fractions: \[ 10a(x^2 - \frac{1}{10}x - \frac{3}{10}) = 10x^2 - x - 3. \] Thus, the polynomial is \( 10x^2 - x - 3 \), which corresponds to option (B). Thus, the correct answer is \( \boxed{10x^2 - x - 3} \).