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If a, b, c are odd integers, then the ro...

If a, b, c are odd integers, then the roots of `ax^(2)+bx+c=0`, if real, cannot be

A

integers

B

rational numbers

C

irrational

D

equal

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To solve the problem, we need to analyze the quadratic equation given by \( ax^2 + bx + c = 0 \) where \( a, b, c \) are odd integers. We want to determine the nature of the roots of this equation when they are real. ### Step-by-Step Solution: 1. **Understanding the Quadratic Equation**: The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients. 2. **Identifying the Conditions**: We know that \( a, b, c \) are odd integers. This means that \( a \equiv 1 \mod 2 \), \( b \equiv 1 \mod 2 \), and \( c \equiv 1 \mod 2 \). 3. **Finding the Discriminant**: The roots of the quadratic equation can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] The term under the square root, \( b^2 - 4ac \), is called the discriminant (\( D \)). For the roots to be real, \( D \) must be non-negative: \[ D = b^2 - 4ac \geq 0 \] 4. **Calculating the Discriminant**: Since \( a, b, c \) are odd, we can express them as: - \( a = 2m + 1 \) - \( b = 2n + 1 \) - \( c = 2p + 1 \) for some integers \( m, n, p \). Now, substituting these into the discriminant: \[ D = (2n + 1)^2 - 4(2m + 1)(2p + 1) \] Expanding this gives: \[ D = 4n^2 + 4n + 1 - 4(4mp + 2m + 2p + 1) \] Simplifying further: \[ D = 4n^2 + 4n + 1 - (16mp + 8m + 8p + 4) \] \[ D = 4n^2 + 4n - 16mp - 8m - 8p - 3 \] 5. **Analyzing the Result**: Since \( D \) is a combination of even and odd terms, we need to see if it can be non-negative. However, notice that \( b^2 \) is odd (since \( b \) is odd) and \( 4ac \) is even (since \( a \) and \( c \) are odd, their product is odd, and multiplying by 4 gives an even number). Thus, \( D \) is odd minus even, which results in an odd number. 6. **Conclusion**: Since \( D \) is odd, it cannot be zero or positive (as it must be non-negative for real roots). Therefore, if the roots are real, they cannot be rational (as rational roots would imply a rational discriminant). ### Final Answer: Thus, the roots of \( ax^2 + bx + c = 0 \), if real, cannot be rational numbers.
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