Chapter 4: Problem 82
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.
Short Answer
Expert verified
(B) \(\geq 0\).
Step by step solution
01
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 \(1 + \frac{c}{a} + \left|\frac{b}{a}\right|\).
02
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 \(\alpha \beta = \frac{c}{a}\).
03
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 \(\frac{c}{a} < -1\).
04
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 \(1 + \text{negative number} + \text{positive number}\).
05
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).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
- Product of the roots: \( \alpha \beta = \frac{c}{a} \)
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.
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:
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.
- The nature and location of the roots.
- The sign of the quadratic expression in 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.