Problem 82 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(a, b, c\) be real, if \(a x^{2}+b x+c=0\) has two real roots \(\alpha\) and \(\beta\), where \(\alpha<-1\) and \(\beta>1\) then \(1+\frac{c}{a}+\left|\frac{b}{a}\right|\) is (A) \(<0\) (B) \(\geq 0\) (C) \(\leq 0\) (D) None of these.

Step-by-Step Solution

Answer

(B) \(\geq 0\).

Step 1: Understand Given Information

We are given a quadratic equation \(a x^{2} + b x + c = 0\) with real roots \(\alpha\) and \(\beta\), where \(\alpha < -1\) and \(\beta > 1\). We need to determine the value of \(1 + \frac{c}{a} + \left|\frac{b}{a}\right|\).

Step 2: Use Vieta's Formulas

According to Vieta's formulas, the sum of the roots \(\alpha + \beta = -\frac{b}{a}\) and the product of the roots \(\alpha \beta = \frac{c}{a}\).

Step 3: Analyze Root Conditions

Given that \(\alpha < -1\) and \(\beta > 1\), their sum \(\alpha + \beta < -1 + 1 = 0\). So \(-\frac{b}{a} < 0\) implying \(\frac{b}{a} > 0\). Similarly, \(\alpha \beta < (-1)(1) = -1\) thus \(\frac{c}{a} < -1\).

Step 4: Evaluate the Expression

The expression is \(1 + \frac{c}{a} + \left|\frac{b}{a}\right|\). Since \(\frac{c}{a} < -1\), we have \(1 + \frac{c}{a} < 0\). Additionally, \(|\frac{b}{a}| = \frac{b}{a}\) because \(\frac{b}{a} > 0\). Thus, \(\left|\frac{b}{a}\right|\) is a positive number. Therefore, the expression becomes \(1 + \text{negative number} + \text{positive number}\).

Step 5: Draw Conclusion

Since \(1 + \frac{c}{a} < 0\) and the addition of a positive number may bring the total to zero or above, the expression \(1 + \frac{c}{a} + \left|\frac{b}{a}\right| \geq 0\). This allows for equality or positivity, thus confirming Option (B).

Key Concepts

Vieta's formulas

Vieta's formulas are a fundamental tool in solving quadratic equations. They offer a direct link between the coefficients of the quadratic equation and its roots. For any quadratic equation of the form \[ a x^2 + b x + c = 0, \]the roots \( \alpha \text{ and } \beta \),can be represented as:

• Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
• Product of the roots: \( \alpha \beta = \frac{c}{a} \)

These relationships help simplify complex algebraic problems by reducing them to more manageable expressions.
Vieta's formulas are not only used to find the roots but also to understand the behavior of the quadratic equation by analyzing the coefficients.

Real roots of quadratic equations

A quadratic equation can have different types of roots depending on its discriminant, \( b^2 - 4ac \).

• If the discriminant is positive, \( b^2 - 4ac > 0 \), the quadratic has two distinct real roots.
• If the discriminant is zero, \( b^2 - 4ac = 0 \), the quadratic has exactly one real root, often called a double root.
• If the discriminant is negative, \( b^2 - 4ac < 0 \), the quadratic does not have real roots; instead, it has complex roots.

Quadratic equations with two distinct real roots are particularly interesting because they define a parabola that intersects the x-axis at two points. Understanding the conditions for real roots allows you to determine the curve's intersection with the x-axis, offering insights into the equation's behavior and graph.

Inequalities in quadratic expressions

Inequalities in quadratic expressions often require a deep understanding of the roots and their relations. When solving quadratic inequalities, key considerations include:

• The nature and location of the roots.
• The sign of the quadratic expression in different intervals.

For example, if you know a quadratic expression has roots such that \( \alpha < -1 \) and \( \beta > 1 \),you can infer that the expression takes specific signs over different intervals.
Sign Analysis: Between the roots, for a quadratic \( ax^2 + bx + c \),where \( a > 0 \),the expression is negative. Outside these roots, it becomes positive.
Such insights are essential when determining solutions to inequalities and verifying expressions involving absolute values, like \( 1 + \frac{c}{a} + \left|\frac{b}{a}\right| \). Understanding how to manipulate these factors helps in solving complex problems involving quadratic inequalities.