Problem 88 - Elementary Algebra

Question

Find a quadratic equation with integer coefficients, given the following solutions. $$ -1 / 2,1 / 2 $$

Step-by-Step Solution

Answer

The quadratic equation is \(4x^2 - 1 = 0\).

Step 1: Understand the Objective

We need to find a quadratic equation with integer coefficients given the solutions \(-\frac{1}{2}\) and \(\frac{1}{2}\). This means the equation should look like \(ax^2 + bx + c = 0\).

Step 2: Form the Factored Equation

Since we know the roots (solutions) of the equation, we can write it in its factored form. If \(\alpha = -\frac{1}{2}\) and \(\beta = \frac{1}{2}\) are the roots, then the factored form of the equation is:\[(x - \alpha)(x - \beta) = 0\] which becomes:\[(x + \frac{1}{2})(x - \frac{1}{2}) = 0\].

Step 3: Use the Difference of Squares Formula

Recognize that the expression \((x + \frac{1}{2})(x - \frac{1}{2})\) is a difference of squares. Apply the formula \(a^2 - b^2 = (a + b)(a - b)\) where \(a = x\) and \(b = \frac{1}{2}\) to simplify the expression:\[x^2 - \left(\frac{1}{2}\right)^2 = x^2 - \frac{1}{4}\].

Step 4: Eliminate the Fraction

To ensure that the equation has integer coefficients, multiply every term by 4 to eliminate the fraction:\[4(x^2 - \frac{1}{4}) = 4x^2 - 1 = 0\].

Step 5: Write the Quadratic Equation

The quadratic equation with integer coefficients that has the given roots \(-\frac{1}{2}\) and \(\frac{1}{2}\) is:\[4x^2 - 1 = 0\].

Key Concepts

Integer Coefficients

When we talk about integer coefficients in a quadratic equation, it means all the numbers in front of the variables are whole numbers—no fractions or decimals allowed. For example, in the equation \(ax^2 + bx + c = 0\), the values \(a\), \(b\), and \(c\) should all be integers. This is a common format because integers are simple and straightforward to work with. In our original exercise, the goal is to maintain integer coefficients, even though the given solutions involve fractions. We'll use math tricks, like multiplying through by a common number, to keep everything integral while still being true to the roots provided initially.

Difference of Squares

The difference of squares is a special algebraic concept where two square terms are subtracted from each other. This formula is written as \(a^2 - b^2 = (a + b)(a - b)\). You spot this pattern whenever you have two terms, such as \(x^2 - \frac{1}{4}\), that fit the profile of this formula.In our step-by-step solution, we identified \((x + \frac{1}{2})(x - \frac{1}{2}) = 0\) as a difference of squares problem. By recognizing that we could apply this formula, it simplifies things down to \(x^2 - \frac{1}{4}\), aligning the given roots more closely with a standard quadratic form. This method is like a mathematical shortcut for dealing with quadratic expressions efficiently.

Factored Form

A quadratic equation's factored form involves expressing it as the product of two linear terms. If you have the roots, also known as solutions, you can write the equation as \((x - \alpha)(x - \beta) = 0\). Here, \(\alpha\) and \(\beta\) represent the roots of the equation.For our exercise, starting with the factored form \((x + \frac{1}{2})(x - \frac{1}{2}) = 0\) aligned perfectly with the given roots \(-\frac{1}{2}\) and \(\frac{1}{2}\). Factoring is beneficial because it's simple to reverse—if you have the factorized form, you can easily find the roots again. This makes it a powerful tool not just for solving, but for crafting equations.

Solutions of Equations

Finding the solutions of a quadratic equation is the same as finding the roots of the equation. This means identifying the values of \(x\) that make the equation true, or cause the whole expression to equal zero.In the solved problem, we started with known solutions. These were \(-\frac{1}{2}\) and \(\frac{1}{2}\). But instead of staying with the fractional form, the next step was to structure the equation such that multiplying through by 4 resulted in integer coefficients. The final equation \(4x^2 - 1 = 0\) confirms those solutions because if you substitute them back into the equation, each turns it into zero, fulfilling the requirements for solutions in a quadratic context.