Problem 47 - Basic Technical Mathematics with Calculus

Question

In Exercises 47 and \(48,\) find the indicated quadratic equations.Find a quadratic equation for which the solutions are 0.5 and 2 .

Step-by-Step Solution

Answer

The quadratic equation is \(x^2 - 2.5x + 1 = 0\).

Step 1: Understand the Problem

In this problem, we are asked to form a quadratic equation that has specific solutions. Our roots or solutions are given as \( x_1 = 0.5 \) and \( x_2 = 2 \). We will use these roots to construct the quadratic equation.

Step 2: Use the Factor Theorem

The Factor Theorem tells us that if \( x_1 \) and \( x_2 \) are roots of the equation, then the factors of the equation are \((x-x_1)\) and \((x-x_2)\). Thus, our initial equation setup is \((x - 0.5)(x - 2) = 0\).

Step 3: Expand The Factors

Now, we expand the factors \((x - 0.5)(x - 2)\) using the distributive property.\[(x - 0.5)(x - 2) = x(x - 2) - 0.5(x - 2) \]Expanding further, we get:\[x^2 - 2x - 0.5x + 1 = 0\]

Step 4: Simplify the Equation

Combine like terms to simplify the expression to form the quadratic equation.\[x^2 - 2.5x + 1 = 0\]This is the quadratic equation whose solutions are \(x = 0.5\) and \(x = 2\).

Key Concepts

Factor Theorem

One of the fundamental concepts in algebra when dealing with quadratic equations is the Factor Theorem. The Factor Theorem connects the roots (or solutions) of a polynomial equation to its factors. This theorem essentially tells us that if a number is a root of a polynomial, the polynomial can be factored using that number.
For a quadratic equation like the one we are given, if your roots are known, say \( x_1 \) and \( x_2 \), you can express the equation as \((x - x_1)(x - x_2) = 0\).

• This theorem is essential because it gives us an easy way to construct polynomial equations when we know their solutions.
• The roots are the values of \( x \) that satisfy the equation (making it equal to zero).
• In our exercise, knowing that \( 0.5 \) and \( 2 \) are roots allowed us to easily express the equation as \((x - 0.5)(x - 2) = 0\).

Using the Factor Theorem streamlines the process of equation creation, turning a list of numbers into a structured mathematical expression.

Roots of Equations

The roots of an equation are the solutions that satisfy the equation, making it equal to zero. In simpler terms, these are the values that make the equation true.
When dealing with quadratic equations, you will generally encounter up to two real roots or solutions.
These are often found by seeing where the graph of the equation crosses the x-axis.

• For the quadratic equation provided, the roots were given as \( x = 0.5 \) and \( x = 2 \).
• This means that if you substitute these values into the equation, each will make it equal to zero.
• The sum and product of the roots are linked by the equation's coefficients in what is known as Viète's formulas: If the polynomial is \( ax^2 + bx + c = 0 \), the sum of the roots is \(-b/a\) and the product is \(c/a\).

Understanding roots helps in both solving equations and in creating them from given solutions.

Algebra

Algebra is a branch of mathematics that deals with variables and the rules for manipulating these variables. It forms the backbone of high school mathematics curriculum, providing tools to solve equations and understand relationships between numbers.
In this context, algebra helps us to craft equations from given pieces of information, like roots.

• Algebraic manipulation allows you to transform expressions and solve equations.
• For instance, in our exercise, using algebraic techniques helped us expand the factors \((x - 0.5)(x - 2)\) into a standard quadratic form.
• By using basic algebraic operations such as distribution, we expressed the multiplication of these two binomials as \(x^2 - 2.5x + 1 = 0\).

Algebraic skills like understanding distribution, combining like terms, and rearranging equations are crucial in mathematics and problem-solving.