Problem 112 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The roots of the equation \(a x^{2}+b x+c=0\), where \(a \in R^{+}\), are two consecutive odd positive integers, then (A) \(|b| \leq 4 a\) (B) \(|b| \geq 4 a\) (C) \(|b|=2 a\) (D) None of these

Step-by-Step Solution

Answer

The roots condition confirms that (B) \(|b| \geq 4a\) is correct.

Step 1: Define Consecutive Odd Integers

Let the roots of the quadratic equation be two consecutive odd integers, say \( m \) and \( m + 2 \). Since they are the roots of the quadratic equation \( ax^2 + bx + c = 0 \), by Vieta's formulas, we find that the sum and product of the roots are related to \( b \) and \( c \).

Step 2: Use Vieta’s Formulas for Sum of Roots

According to Vieta’s formulas, the sum of the roots \( m + (m + 2) = 2m + 2 \) equals \(-\frac{b}{a}\). Therefore, \( -\frac{b}{a} = 2m + 2 \), which gives us \( b = -a(2m + 2) \).

Step 3: Derive the Condition for |b|

The expression \( b = -a(2m + 2) \) implies that \(|b| = |a(2m + 2)| = 2a(m+1)\). Since \( m \) is an odd integer, \( m + 1 \) is an even integer, and the smallest possible value for \( m + 1 \) when \( m \) is a positive odd integer is 2 (when \( m = 1 \)). This results in \(|b| = 4a\).

Step 4: Conclusion Based on Derived Inequality

Given that the smallest possible value of \(|b|\) is \( 4a \) for any positive integer \( m \), we can definitively determine that \(|b| \geq 4a\). This implies that option (B) \(|b| \geq 4a\) is the correct answer.

Key Concepts

Consecutive Odd Integers

Consecutive odd integers are numbers that come one after the other in the sequence of odd numbers. For example, if you start with any odd number like 3, the next odd number would be 5, and so on, following the pattern of adding 2. This means that if you have one consecutive odd integer, the next one can simply be found by adding 2 to it.

When dealing with problems like the one in the exercise, it's helpful to label these integers as \( m \) and \( m + 2 \). These labels allow us to set up equations and solve problems systematically. Using algebra to define these integers assists in connecting them to the roots of a quadratic equation.

Vieta's Formulas

Vieta's formulas are a fascinating way to link the coefficients of a polynomial to the sums and products of its roots. For a quadratic equation like \( ax^2 + bx + c = 0 \), these formulas become particularly useful.

They tell us:

• The sum of the roots \( x_1 + x_2 = -\frac{b}{a} \)
• The product of the roots \( x_1 \cdot x_2 = \frac{c}{a} \)

In our context, we used the sum of roots part of Vieta's formulas to relate \( 2m + 2 \) (sum of our consecutive integers roots) to \(-\frac{b}{a}\). This relationship helps us express \( b \) in terms of \( a \) and \( m \). It's one of those key tools in algebra that offers insights into the characteristics of polynomial roots without having to solve the entire equation directly.

Roots of a Quadratic

In the realm of quadratic equations, the term 'roots' refers to the solutions of the equation \( ax^2 + bx + c = 0 \). These roots are the values of \( x \) that would make the equation true, essentially where the graph of the quadratic function crosses the x-axis.

Quadratic equations can have two real roots, one real root, or no real roots depending on the discriminant (\( b^2 - 4ac \)). For problems involving integers, and particularly here with consecutive odd integers, we're looking for specific integer values that solve the equation.

In this scenario, by setting the roots as \( m \) and \( m + 2 \), we are essentially stating that both values are solutions to the quadratic equation. This allows us to derive relationships between the equation's coefficients based on Vieta's formulas. It forms the base on which we apply additional algebraic steps to find necessary conditions such as those relating to \( b \).