Problem 50 - College Algebra

Question

Show that the sum of the two solutions to the quadratic equation is \(-\frac{b}{a}\) .

Step-by-Step Solution

Answer

The sum of the solutions is \(-\frac{b}{a}\).

Step 1: Identify the Quadratic Equation

A quadratic equation is generally given in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\).

Step 2: Recall Vieta's Formula

According to Vieta's formulas, the sum of the roots (solutions) for the quadratic equation \(ax^2 + bx + c = 0\) is given by \(-\frac{b}{a}\), and the product of the roots is \(\frac{c}{a}\).

Step 3: Establish the Roots of the Quadratic Equation

The roots of the quadratic equation can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 4: Calculate the Sum of the Roots

Using the quadratic formula results, calculate the sum of the two solutions: \(x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a}\).

Step 5: Simplify the Expression for Sum of Roots

Combine the numerators over a common denominator: \(x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4ac} + (-b) - \sqrt{b^2 - 4ac}}{2a}\). The \(\sqrt{b^2 - 4ac}\) terms cancel out, leaving: \(x_1 + x_2 = \frac{-2b}{2a}\).

Step 6: Final Simplification

Simplify the result: \(x_1 + x_2 = -\frac{b}{a}\). This confirms Vieta’s formula and shows that the sum of the solutions is indeed \(-\frac{b}{a}\).

Key Concepts

Vieta's Formulas

Vieta's formulas provide a fascinating bridge between the coefficients of a quadratic equation and its roots, making it simpler to understand some key properties.
For the quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's formulas offer two major insights:

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

These insights are incredibly useful for quickly calculating the nature of roots without needing to solve the entire equation. By analyzing the coefficients directly, Vieta's formulas save time and offer a deeper understanding of quadratic relationships. Let's not forget that these formulas apply beautifully to real numbers and complex numbers, offering a broad application. This makes them a handy tool in both theoretical math and practical applications.

Quadratic Formula

The quadratic formula is an essential tool for finding the roots of any quadratic equation. It's a reliable method derived from the process of completing the square. Here's the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In this formula, the expression under the square root \(b^2 - 4ac\), known as the discriminant, determines the nature of the roots:

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root (a repeated root).
• If \(b^2 - 4ac < 0\), the roots are complex (conjugate pairs).

This tool is indispensable for solving all forms of quadratic equations, providing precise answers both for professionals and students. Remember, while using the quadratic formula, careful attention needs to be given to sign changes and arithmetic to get accurate results.

Roots of a Quadratic Equation

The roots of a quadratic equation are the solutions \(x_1\) and \(x_2\) that satisfy \(ax^2 + bx + c = 0\). These roots can be obtained through various methods, but the quadratic formula is often preferred for its reliability and straightforwardness.
To find these roots using the quadratic formula:1. Calculate the discriminant \( (b^2 - 4ac) \) and determine the nature of the roots.2. Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula to get the exact roots. 3. Simplify the expression to resolve the values for \(x_1\) and \(x_2\).
The process not only provides the roots but also gives insight into the parabola's graph characteristics. For example, knowing whether the roots are real or complex affects the graph's intersections with the x-axis. Root determination is fundamental for wider applications, including physics and engineering, where quadratic equations are frequently encountered.