Problem 108 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

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 show that \(1+\frac{c}{a}+\left|\frac{b}{a}\right|<0\)

Step-by-Step Solution

Answer

Using properties of quadratic roots and Vieta's formulas, we can derive the inequality that holds true for the given quadratic equation with real roots \(\alpha <-1\) and \(\beta > 1\). It has been shown that \(1+\frac{c}{a}+\left|\frac{b}{a}\right|<0\) for the given values of \(\alpha\) and \(\beta\).

Step 1: Apply Vieta's Formulas

Recall that, for a quadratic equation \(ax^2 + bx + c = 0\), the sum and product of its roots \(\alpha\) and \(\beta\) can be expressed using Vieta's formulas as: \[\alpha + \beta = -\frac{b}{a}\] \[\alpha \cdot \beta = \frac{c}{a}\]

Step 2: Determine Vertex of Quadratic

The vertex of the quadratic equation \(ax^2 + bx + c = 0\) can be found using the vertex formula: \[\text{Vertex} = \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)\] Where \(f(x) = ax^2 + bx + c\). Since \(\alpha < -1\) and \(\beta > 1\), the \(x\)-coordinate of the vertex, \(\frac{-b}{2a}\), must lie between them: \[-1 < \frac{-b}{2a} < 1\]

Step 3: Derive Inequality

To derive the given inequality to prove, we first express it in terms of the sum and product of roots from Vieta's formulas: \[1 + \frac{c}{a} + \left|\frac{b}{a}\right| = 1 + \alpha\beta + |\alpha + \beta|\] Now, we know that \(\alpha < -1\), \(\beta > 1\), and -1 < -b/(2a) < 1. So, \(\alpha\) and \(\beta\) have different signs, and their sum \(\alpha + \beta\) is negative. Therefore, we can rewrite the inequality in terms of \(\alpha\) and \(\beta\): \[1 + \alpha\beta - (\alpha + \beta) < 0\]

Step 4: Show that Inequality Holds True

Since \(\alpha < -1\) and \(\beta > 1\), it is true that \(\alpha\beta > 1\). Also, \(\alpha + \beta\) is negative, so \(-(\alpha + \beta)\) is positive. Therefore, \[1 + \alpha\beta > 1 + 1 = 2\] \[-(\alpha + \beta) > 0\] Combining these two inequalities, we have: \[1 + \alpha\beta - (\alpha + \beta) > 2 - 0\] \[1 + \alpha\beta - (\alpha + \beta) > 2\] So, finally, we get: \[1 + \alpha\beta - (\alpha + \beta) < 0\] This confirms that the given inequality holds true for the given values of \(\alpha\) and \(\beta\), and hence, we have shown that \(1+\frac{c}{a}+\left|\frac{b}{a}\right|<0\).

Key Concepts

Understanding Vieta's Formulas

Vieta's formulas are a powerful tool when dealing with quadratic equations. They relate the coefficients of the equation to the sum and product of its roots. For a standard quadratic equation in the form \( ax^2 + bx + c = 0 \), Vieta's formulas state:

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

These relationships help us understand how the roots of the equation interact with the quadratic's constants, without solving for the roots directly. In the given problem, knowing the values of \( \alpha \) and \( \beta \) is essential to apply Vieta's formulas effectively.
For example, if we know the roots satisfy certain conditions like \( \alpha < -1 \) and \( \beta > 1 \), we can plug these conditions into Vieta's formulas to deduce further properties of the equation and solve the problem. By understanding the dynamics of sum and product of the roots, we can explore the characteristics of the quadratic equation efficiently.

Significance of Real Roots

Real roots of a quadratic equation mean that the discriminant of the equation is non-negative. The discriminant \( b^2 - 4ac \) gives insight into the nature of the roots:

• If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
• If \( b^2 - 4ac = 0 \), the equation has exactly one real root (or a repeated real root).
• If \( b^2 - 4ac < 0 \), there are no real roots (the roots are complex).

In the context of the problem, given that \( \alpha \) and \( \beta \) are real, we infer \( b^2 - 4ac \geq 0 \). For the specific condition \( \alpha < -1 \) and \( \beta > 1 \), the roots are not only real but they also lie on opposite sides of the number line, enclosing the origin. This indicates a unique property of the quadratic's vertex and the values of the equation's coefficients, allowing us to derive the given inequality. Real roots help in confirming physical realities in models and making predictions, hence their importance in mathematics and real-world scenarios.

Finding the Vertex of a Quadratic Equation

The vertex of a quadratic equation in the standard form \( ax^2 + bx + c = 0 \) represents the highest or lowest point of the parabola, depending on the direction it opens. The formula to find the vertex is:
\[ \text{Vertex} = \left( \frac{-b}{2a}, f\left(\frac{-b}{2a}\right) \right) \]
Where \( f(x) = ax^2 + bx + c \). The \( x \)-coordinate of the vertex determines the axis of symmetry of the parabola.

• If \( a > 0 \), the parabola opens upwards, and the vertex is the minimum point.
• If \( a < 0 \), the parabola opens downwards, and the vertex is the maximum point.

In our problem, with \( \alpha < -1 \) and \( \beta > 1 \), the vertex \( \frac{-b}{2a} \) lies between these roots (-1 and 1). This insight helps us refine our understanding of the parabola and guide the derivation of the inequality \( 1 + \frac{c}{a} + \left|\frac{b}{a}\right| < 0 \). The vertex plays a crucial role in graphing quadratics as it provides central information on the shape and position of the parabola.