If a, b, c are odd integers and aX^2 + bX +c =0 has real roots, then (a) both roots are rational(b) both roots are irrational (c) both roots are positive (d) roots are of opposite sign

To determine the nature of the roots of the quadratic equation \(aX^2 + bX + c = 0\) where \(a\), \(b\), and \(c\) are odd integers, we first need to consider the discriminant of the equation. The discriminant, denoted as \(D\), is given by the formula \(D = b^2 - 4ac\). For the roots to be real, the discriminant must be non-negative, meaning \(D \geq 0\).

Analyzing the Discriminant

Let’s analyze the components of the discriminant with \(a\), \(b\), and \(c\) as odd integers:

• Odd Integer Properties: The square of any odd integer is also odd. Therefore, \(b^2\) will be odd.
• Product of Odd Integers: The product \(4ac\) involves multiplying \(4\) (an even integer) with \(a\) and \(c\) (both odd integers). This results in \(4ac\) being even.

Since \(D = b^2 - 4ac\), we can conclude that \(b^2\) is odd and \(4ac\) is even. When we subtract an even number from an odd number, the result is always odd. Thus, \(D\) must be odd, which means \(D\) is strictly greater than zero.

Implications of a Positive Discriminant

With \(D > 0\), we can deduce that there are two distinct real roots. However, we need to delve deeper into whether these roots are rational or irrational.

The roots of the quadratic equation can be expressed using the quadratic formula:

X = \frac{-b \pm \sqrt{D}}{2a}

Examining the Roots

Now, since \(D\) is odd, \(\sqrt{D}\) will be irrational unless \(D\) is a perfect square. However, since we are dealing with odd integers, \(D = b^2 - 4ac\) is unlikely to be a perfect square. Thus, it is highly probable that \(\sqrt{D}\) is irrational. This leads us to conclude that the roots themselves will be irrational because:

-b is even (since b is odd) and 2a is even (since a is odd).

Thus, the entire expression \(\frac{-b \pm \sqrt{D}}{2a}\) will yield irrational results due to the presence of the irrational square root.

Root Characteristics

Next, we should consider the signs of the roots. The nature of the roots being irrational implies they could be either both positive or of opposite signs, but let’s analyze this using the properties of quadratic functions:

Since \(a\) is positive (as it is an odd integer), the parabola opens upwards. For a quadratic function that opens upwards with a positive discriminant, if the roots are irrational and \(D > 0\), there’s a significant chance that one root lies below zero while the other is above zero. Thus, it can be inferred that the roots are of opposite signs.

Final Assessment

To summarize:

• Both roots are irrational.
• Both roots are likely to be of opposite signs due to the nature of the quadratic function.

Thus, the answer to the question posed is that the roots are of opposite sign. This leads us to select option (d) as the correct choice.