Problem 9 - Algebra for College Students

Question

For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions. See Example 1. $$\sqrt{5},-\sqrt{5}$$

Step-by-Step Solution

Answer

The quadratic equation is \( x^2 - 5 = 0\).

Step 1: Understand the Problem

We are given a pair of numbers and need to create a quadratic equation with integral coefficients that has these numbers as its solutions. The given numbers are \(\begin{array}{c}\sqrt{5}, \ -\sqrt{5}\end{array}\).

Step 2: Recall the Form of a Quadratic Equation

A quadratic equation with roots in the form of \( \alpha \ and \ \beta \) can be written as \((x - \alpha) (x - \beta) = 0\).

Step 3: Set Up the Equation Using the Given Roots

Substitute \( \sqrt{5} \ and \ - \sqrt{5} \) into the quadratic equation form: \((x - \sqrt{5})(x + \sqrt{5}) = 0\).

Step 4: Expand the Equation

Expand the product \( (x - \sqrt{5})(x + \sqrt{5})\) using the difference of squares formula: \( x^2 - (\sqrt{5})^2 = 0\).

Step 5: Simplify the Equation

Simplify \( x^2 - (\sqrt{5})^2\) to \( x^2 - 5 = 0\).

Key Concepts

Integral Coefficients

A quadratic equation is an equation of the form: \[ax^2 + bx + c = 0\] where a, b, and c are constants, and the form of this equation ensures that its degree is 2. For a quadratic equation to have integral coefficients, all the constants \(a, b,\) and \(c\) must be whole numbers.
This is important in many mathematical problems because it often simplifies calculations and ensures that the solutions or roots of the equation are rational numbers. For example, in our step-by-step solution, when we expand \((x - \sqrt{5})(x + \sqrt{5})\) and simplify it to \(x^2 - 5 = 0\), all the coefficients, \(1\) for \(x^2\) and \(-5\) for the constant term, are integers.
By ensuring integral coefficients, we make the equation easier to handle and interpret. This is exactly what we achieved in the problem where the given roots \( \sqrt{5}\) and \(-\sqrt{5}\) led to the simplified quadratic form \(x^2 - 5 = 0\).

Roots of a Quadratic Equation

The roots of a quadratic equation \(ax^2 + bx + c = 0\) are values of \(x\) that satisfy this equation.
These values are also known as the solutions to the quadratic equation. There are a few different methods to find these roots: factorizations, graphing, completing the square, and using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our example, the roots were provided directly as \( \sqrt{5}\) and \(-\sqrt{5}\). To find a quadratic equation with given roots \( \alpha \) and \( \beta \), you can use the fact that the quadratic equation can be written in the form \((x - \alpha)(x - \beta) = 0\).
For the given problem, substituting \( \sqrt{5} \) and \( -\sqrt{5} \) into \((x - \alpha)(x - \beta)\) yields \((x - \sqrt{5})(x + \sqrt{5}) = 0\). Expanding this, we get \( x^2 - 5 = 0\), as illustrated in the step-by-step solution.
This approach is often very useful when dealing with roots that are not easily manageable numbers.

Difference of Squares

The difference of squares is a special case of polynomial multiplication, where you multiply a binomial by its conjugate. The general form is \[a^2 - b^2 = (a - b)(a + b)\]
The difference of squares formula can greatly simplify the multiplication process for certain pairs of terms. When looking at our problem, the roots provided were \(\sqrt{5}\) and \(-\sqrt{5}\). Substituting these values into the binomial form, we get \[(x - \sqrt{5})(x + \sqrt{5}) = 0\]
When expanded, this results in: \[x^2 - (\sqrt{5})^2 = 0\] which further simplifies to: \[x^2 - 5 = 0\]
Notice how we used the difference of squares to manage the square root terms and transform them into simple whole numbers in the quadratic equation. This property helps in simplifying complex algebraic expressions and is especially useful in problems involving roots of naturally occurring numbers, such as square roots.