Problem 109 - Algebra and Trigonometry Real Mathematics, Real People

Question

Given that the solutions of a quadratic equation are \(x=(-b \pm \sqrt{b^{2}-4 a c}) /(2 a),\) show that the sum of the solutions is \(S=-b / a\).

Step-by-Step Solution

Answer

The sum of the solutions \(S=x_{1}+x_{2}=-b / a\)

Step 1: Identify the solutions of the quadratic equation

The quadratic formula for solving equations in the form \(ax^{2}+bx+c=0\) is \(x=(-b \pm \sqrt{b^{2}-4ac}) /(2a)\). According to this formula, the solutions of the quadratic equation are \(x_{1}=(-b + \sqrt{b^{2}-4ac}) /(2a)\) and \(x_{2}=(-b - \sqrt{b^{2}-4ac}) /(2a)\).

Step 2: Compute the sum of the solutions

Sum of solutions, denoted as S, is given by \(S=x_{1}+x_{2}\). If we substitute the values of \(x_{1}\) and \(x_{2}\), we get \(S=(-b + \sqrt{b^{2}-4ac}) /(2a) + (-b - \sqrt{b^{2}-4ac}) /(2a)\).

Step 3: Simplify the sum of the solutions

On simplification, \(S=(-2b) /(2a) = -b/a\)

Key Concepts

Quadratic Formula

When solving quadratic equations, the quadratic formula is your ultimate tool. It applies to any quadratic equation in the form of \( ax^2 + bx + c = 0 \). Using the quadratic formula helps you find the solutions, or roots, of the equation without the need to factor or complete the square.
The formula looks like this:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]Where:

• \( a \), \( b \), and \( c \) are coefficients of the equation.
• The term \( \pm \) means there are two possible solutions.
• The expression under the square root, \( b^2 - 4ac \), is called the discriminant.

Whether the discriminant is positive, negative, or zero will tell you about the nature of the roots, providing a complete picture of the solutions.
Each root is calculated by considering both plus and minus sides of the formula. This double possibility is what allows the formula to offer up two solutions, \( x_1 \) and \( x_2 \).

Sum of Solutions

The sum of solutions for any quadratic equation can be found using a property derived from the quadratic formula. If you have a quadratic equation \( ax^2 + bx + c = 0 \), the sum of its solutions, \( S \), is remarkably straightforward to calculate.
From the equations \( x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \) and \( x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \), adding these two solutions will always lead you to \( S = -\frac{b}{a} \).
Let's see why:

• When you add \( x_1 \) and \( x_2 \), the terms \( \sqrt{b^2 - 4ac} \) and \( -\sqrt{b^2 - 4ac} \) cancel each other out.
• What's left is \( -b/a \), since you simply sum up \( -b \) and divide by \( a \).

This relationship is deeply rooted in the structure of quadratic equations and helps in predicting the nature of the solutions before finding them individually.

Polynomial Equation Solving

Solving any polynomial equation, particularly quadratic ones, seeks to find the values of \( x \) that make the equation true. With quadratics, the consistent method of finding roots is through the quadratic formula.
Quadratic equations are typically the only degree-two polynomials. They always form a parabola graphically, opening either upwards or downwards based on the sign of \( a \). The solutions, or roots, of the polynomial mark where the graph intersects the x-axis.
There are several methods to solve these equations:

• **Factoring:** Works when the quadratic expression can be expressed as the product of two binomials.
• **Completing the Square:** Converts the equation to the form \( (x - p)^2 = q \), making it simpler to solve.
• **Quadratic Formula:** The universal approach that works regardless of whether the quadratic can be easily factored or not.

Applying these techniques allows you to determine the values of \( x \) easily, providing a deeper understanding of the behavior of quadratic functions.