Problem 78 - A Graphical Approach to Precalculus with Limits

Question

For each pair of numbers, find the values of \(a, b,\) and \(c\) for which the quadratic equation ax \(^{2}+b x+c=0\) has the given numbers as solutions. Answers may vary. (Hint: Use the zero-product property in reverses $$-3,2$$

Step-by-Step Solution

Answer

The values are \(a = 1\), \(b = 1\), \(c = -6\).

Step 1: Understand the Problem

We are given two solutions for a quadratic equation, \(-3\) and \(2\). We need to find the coefficients \(a\), \(b\), and \(c\) of the quadratic equation \(ax^2 + bx + c = 0\) that has these roots.

Step 2: Use the Zero-Product Property

The equation can be derived using the zero-product property in reverse. If \(x_1\) and \(x_2\) are the roots, then the quadratic equation can be expressed as \((x - x_1)(x - x_2) = 0\). In this case, \((x + 3)(x - 2) = 0\).

Step 3: Expand the Factored Form

Expand \((x + 3)(x - 2) = 0\) to find the quadratic expression:\[(x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6\]

Step 4: Identify the Coefficients

From the expanded form \(x^2 + x - 6\), compare it with the standard form \(ax^2 + bx + c = 0\): - \(a = 1\)- \(b = 1\)- \(c = -6\)

Step 5: Verify the Solution

To verify, double-check the calculations:- Substitute \(x = -3\): \((-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0\)- Substitute \(x = 2\): \(2^2 + 2 - 6 = 4 + 2 - 6 = 0\)Both calculations give zero, confirming the roots.

Key Concepts

Zero-Product Property

The Zero-Product Property is a fundamental concept in algebra that tells us when the product of two things is zero, at least one of those things must be zero. It is particularly useful when solving quadratic equations in factored form.

To understand how this works, think of the equation \( (x - x_1)(x - x_2) = 0 \). Here, \(x_1\) and \(x_2\) are the solutions or roots of the quadratic equation. The Zero-Product Property implies that either \(x - x_1 = 0\) or \(x - x_2 = 0\), and thus \(x = x_1\) or \(x = x_2\).

In our solved problem, the given solutions for the quadratic equation were \(-3\) and \(2\). By using the Zero-Product Property, we expressed the equation in factored form as \((x + 3)(x - 2) = 0\). Each factor corresponds to one of the solutions, effectively working backwards to construct the quadratic equation from its known roots.

Factored Form

In the context of quadratic equations, the factored form is a representation that reveals the roots or solutions of the equation directly. The general form of a quadratic equation is \(ax^2 + bx + c = 0\), but when it's in factored form, it appears as \((x - x_1)(x - x_2) = 0\).

To achieve the factored form, we use reverse engineering from the solutions or roots of the equation. Once we know the roots, we can write the equation as \((x - x_1)(x - x_2)\).

In our exercise, we had solutions \(-3\) and \(2\), which allowed us to write the equation in factored form as \((x + 3)(x - 2) = 0\). This method gives a quick and easy way to express the relationship between a quadratic equation and its roots. Once expanded to the general form, we obtained \(x^2 + x - 6 = 0\), helping us to clearly identify the coefficients \(a = 1, b = 1, c = -6\).

Roots of Quadratic Functions

Roots or solutions of quadratic functions are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). More intuitively, they are the points where the graph of the quadratic function intersects the \(x\)-axis.

In basic terms, if a quadratic function is factored as \((x - x_1)(x - x_2) = 0\), then \(x_1\) and \(x_2\) are considered the roots, because they make each respective part of the product zero. These roots are valuable since they provide insights into the properties of the quadratic function, like its direction, width, and location on the Cartesian plane.

For our problem, the roots were \(-3\) and \(2\). By knowing these, we could easily create the factored form of the quadratic equation, and further derive the standard quadratic form. The roots were verified by plugging back into the equation \(x^2 + x - 6 = 0\) and ensuring both values satisfied the condition, confirming the correctness of our derived equation.