Two Real Roots Condition

Introduction

Key Concepts

Quadratic Equations and Their Standard Form

A quadratic equation is a second-degree polynomial equation in a single variable $x$, with the standard form: $$ ax^2 + bx + c = 0 $$ where $a$, $b$, and $c$ are coefficients, and $a \neq 0$. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of the coefficient $a$.

Roots of a Quadratic Equation

The solutions to the quadratic equation, known as roots, are the values of $x$ that satisfy the equation. These roots can be real or complex numbers. The nature of the roots depends on the discriminant, which is derived from the coefficients of the equation.

Discriminant and Its Role

The discriminant ($\Delta$) of a quadratic equation is given by the formula: $$ \Delta = b^2 - 4ac $$ The discriminant determines the nature of the roots:

• If $\Delta > 0$, the equation has two distinct real roots.
• If $\Delta = 0$, the equation has exactly one real root (a repeated root).
• If $\Delta

Two Real Roots Condition

For a quadratic equation to have two real roots, the discriminant must be positive. This condition ensures that the parabola intersects the x-axis at two distinct points. Mathematically, the condition is expressed as: $$ b^2 - 4ac > 0 $$ This inequality must be satisfied for the equation to possess two real solutions.

Deriving the Two Real Roots Condition

Starting with the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ The term under the square root, $\sqrt{b^2 - 4ac}$, determines the nature of the roots. For the roots to be real and distinct, the expression inside the square root must be positive: $$ b^2 - 4ac > 0 $$ This ensures that the square root is a real number, leading to two different values of $x$.

Graphical Interpretation

Graphically, the two real roots of a quadratic equation correspond to the points where the parabola intersects the x-axis. A positive discriminant implies that the parabola crosses the x-axis at two distinct points, indicating two real solutions.

Examples Illustrating Two Real Roots Condition

Example 1: Consider the quadratic equation $x^2 - 5x + 6 = 0$. Here, $a = 1$, $b = -5$, and $c = 6$. $$ \Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0 $$ Since $\Delta > 0$, the equation has two real roots: $$ x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2} \Rightarrow x = 3 \text{ and } x = 2 $$ Example 2: Consider the quadratic equation $2x^2 + 4x + 2 = 0$. Here, $a = 2$, $b = 4$, and $c = 2$. $$ \Delta = 4^2 - 4(2)(2) = 16 - 16 = 0 $$ Since $\Delta = 0$, the equation has one real repeated root: $$ x = \frac{-4}{4} = -1 $$ Example 3: Consider the quadratic equation $x^2 + 2x + 5 = 0$. Here, $a = 1$, $b = 2$, and $c = 5$. $$ \Delta = 2^2 - 4(1)(5) = 4 - 20 = -16

Applications of Two Real Roots Condition

The two real roots condition is not only a theoretical concept but also has practical applications in various fields such as physics, engineering, and economics. For instance:

• Physics: Determining the points of intersection in projectile motion.
• Engineering: Analyzing the stresses and strains in materials.
• Economics: Modeling profit functions to find break-even points.

Understanding this condition helps in solving real-world problems that can be modeled using quadratic equations.

Common Mistakes to Avoid

When determining the number of real roots, students often make the following mistakes:

• Incorrect calculation of the discriminant, leading to wrong conclusions about the nature of roots.
• Confusing the conditions for one real root and two real roots.
• Neglecting to consider the coefficient $a$ when simplifying the quadratic equation.

It's crucial to carefully compute the discriminant and understand its implications to accurately determine the nature of the roots.

Practice Problems

Problem 1: Determine the nature of the roots for the equation $3x^2 - 2x + 4 = 0$.
Solution: $$ \Delta = (-2)^2 - 4(3)(4) = 4 - 48 = -44 Problem 2: Find the roots of the equation $x^2 - 4x + 4 = 0$.
Solution: $$ \Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0 $$ The equation has one real repeated root: $$ x = \frac{4}{2} = 2 $$ Problem 3: For what values of $k$ does the equation $x^2 + kx + 9 = 0$ have two real roots?
Solution: $$ \Delta = k^2 - 4(1)(9) > 0 \Rightarrow k^2 > 36 \Rightarrow k 6 $$ Thus, the equation has two real roots when $k 6$.

Advanced Concepts

Theoretical Derivations and Proofs

To delve deeper into the two real roots condition, let's explore the derivation of the quadratic formula and its connection to the discriminant. Starting with the standard quadratic equation: $$ ax^2 + bx + c = 0 $$ We can solve for $x$ using the method of completing the square: \begin{align*} ax^2 + bx &= -c \\ x^2 + \frac{b}{a}x &= -\frac{c}{a} \\ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 &= -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*} This derivation highlights the role of the discriminant ($\Delta = b^2 - 4ac$) in determining the nature of the roots. The square root term must be a real number for the roots to be real, which occurs when $\Delta > 0$.

Algebraic Proof of Two Real Roots Condition

Consider the function $f(x) = ax^2 + bx + c$. The nature of its roots is determined by the critical points of the function. 1. **Vertex of the Parabola:** The vertex of the parabola represented by $f(x)$ is at: $$ x = -\frac{b}{2a} $$ Substituting this back into $f(x)$ gives the y-coordinate of the vertex: $$ f\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a} $$ 2. **Condition for Two Real Roots:** For $f(x)$ to intersect the x-axis at two distinct points, the vertex must be below the x-axis (if $a > 0$) or above the x-axis (if $a 0 $$ or $$ c - \frac{b^2}{4a} > 0 \quad \text{if } a 0 $$ Thus, proving that the discriminant must be positive for the quadratic equation to have two real roots.

Complex Problem-Solving

Problem 4: Prove that if a quadratic equation $ax^2 + bx + c = 0$ has two real roots, then the product of the roots is $\frac{c}{a}$ and their sum is $-\frac{b}{a}$. Solution: Let the roots be $x_1$ and $x_2$. According to Vieta's formulas: \begin{align*} x_1 + x_2 &= -\frac{b}{a} \\ x_1 \cdot x_2 &= \frac{c}{a} \end{align*} Given that the equation has two real roots, the discriminant $\Delta = b^2 - 4ac > 0$ ensures that $x_1$ and $x_2$ are real and distinct, validating Vieta's formulas.

Integration with Other Mathematical Concepts

The two real roots condition intersects with various mathematical areas:

• Calculus: Analyzing the extrema of quadratic functions and determining intervals of increase or decrease.
• Linear Algebra: Understanding eigenvalues in the context of quadratic forms.
• Geometry: Solving geometric problems involving areas and distances that lead to quadratic equations.

For example, finding the maximum height of a projectile involves quadratic equations where the discriminant indicates whether the projectile reaches a certain height.

Applications in Real-World Scenarios

Beyond theoretical mathematics, the two real roots condition has practical applications:

• Engineering: Designing structures that require solving quadratic equations for load distribution.
• Finance: Calculating investment returns where profit functions are quadratic.
• Environmental Science: Modeling population growth or decay using quadratic models.

Understanding when quadratic models yield two distinct solutions is crucial for accurate predictions and effective problem-solving in these fields.

Advanced Example:

Problem 5: A company manufactures and sells $x$ units of a product. The cost function is given by $C(x) = 500 + 50x$, and the revenue function is $R(x) = 150x - x^2$. Determine the number of units sold for which the company breaks even, and verify if there are two real solutions. Solution: Break-even occurs when $C(x) = R(x)$: $$ 500 + 50x = 150x - x^2 $$ Rearranging the equation: $$ x^2 - 100x + 500 = 0 $$ Here, $a = 1$, $b = -100$, and $c = 500$. Calculating the discriminant: $$ \Delta = (-100)^2 - 4(1)(500) = 10000 - 2000 = 8000 > 0 $$ Since $\Delta > 0$, there are two real roots. Solving for $x$: $$ x = \frac{100 \pm \sqrt{8000}}{2} = \frac{100 \pm 89.44}{2} $$ Thus, the break-even points are at: $$ x = \frac{100 + 89.44}{2} = 94.72 \quad \text{and} \quad x = \frac{100 - 89.44}{2} = 5.28 $$ Therefore, the company breaks even at approximately 5 and 95 units sold.

Graphical Analysis of Advanced Concepts

Plotting the cost and revenue functions provides a visual representation of the break-even points:

• The cost function $C(x) = 500 + 50x$ is a straight line with a y-intercept of 500 and a slope of 50.
• The revenue function $R(x) = 150x - x^2$ is a downward-opening parabola with its vertex at $x = 75$.

The intersection points of these graphs correspond to the real roots of the equation $C(x) = R(x)$, confirming the existence of two real solutions as derived algebraically.

Exploring the Impact of Coefficients on Roots

The coefficients $a$, $b$, and $c$ in a quadratic equation significantly influence the nature and position of the roots. By manipulating these coefficients, one can control the discriminant and thereby the type of roots:

• Increasing $b$ while keeping $a$ and $c$ constant increases the discriminant, potentially leading to two distinct real roots.
• Adjusting $c$ affects the y-intercept of the graph, influencing the intersection points with the x-axis.
• Changing $a$ alters the width and direction of the parabola, impacting the discriminant indirectly through the $4ac$ term.

Understanding these relationships allows for targeted modifications in applications such as optimization problems and modeling scenarios.

Comparison Table

Aspect	Two Real Roots	One Real Root	No Real Roots
Discriminant ($\Delta$)	$\Delta > 0$	$\Delta = 0$	$\Delta
Number of Roots	Two distinct real roots	One real repeated root	Two complex conjugate roots
Graphical Representation	Parabola intersects x-axis at two points	Parabola tangents the x-axis at the vertex	Parabola does not intersect the x-axis
Example	$x^2 - 5x + 6 = 0$	$x^2 - 4x + 4 = 0$	$x^2 + 2x + 5 = 0$

Summary and Key Takeaways