Problem 95 - Precalculus Student Solutions Manual 5th

Question

Find the values of \(a, b,\) and \(c\) for which the quadratic equation $$ a x^{2}+b x+c=0 $$ has the given numbers as solutions. (Hint: Use the zero-factor property in reverse.) $$4,5$$

Step-by-Step Solution

Answer

The values are \( a = 1 \), \( b = -9 \), and \( c = 20 \).

Step 1: Understand the Problem

Given the quadratic equation \( ax^2 + bx + c = 0 \) and its solutions \( 4 \) and \( 5 \), use the fact that if \( p \) and \( q \) are solutions of a quadratic equation, then the equation can be written in the form \( a(x - p)(x - q) = 0 \).

Step 2: Apply the Zero-Factor Property

Since the solutions are \( 4 \) and \( 5 \), we can write the equation as \( a(x - 4)(x - 5) = 0 \).

Step 3: Expand the Factored Form

Expand \( (x - 4)(x - 5) \) to get the standard quadratic form: \( (x - 4)(x - 5) = x^2 - 9x + 20 \). Therefore, \( a(x^2 - 9x + 20) = 0 \) which simplifies to \( ax^2 - 9ax + 20a = 0 \).

Step 4: Identify the Coefficients

Match the coefficients with the general quadratic form \( ax^2 + bx + c = 0 \): \( a = 1 \), \( b = -9 \), and \( c = 20 \).

Key Concepts

Zero-Factor Property

The Zero-Factor Property is a fundamental principle used to solve quadratic equations. It states that if you have a product of factors equal to zero, then at least one of the factors must be zero. In mathematical terms, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \) (or both).

This property helps us break down complex quadratic equations into simpler linear factors, making it easier to find the solutions. In this particular exercise, because the solutions are given as \( 4 \) and \( 5 \), we can quickly rearrange the quadratic equation into the form \( a(x - 4)(x - 5) = 0 \). Applying the Zero-Factor Property then tells us that either \( x - 4 = 0 \) or \( x - 5 = 0 \), leading directly to the solutions \( x = 4 \) and \( x = 5 \).

Factoring Quadratics

Factoring is a critical skill in solving quadratic equations. It involves expressing the quadratic as a product of two binomials. For instance, if we start from the quadratic form \( a(x-p)(x-q) = 0 \), where \( p \) and \( q \) are the solutions, we can reverse-engineer the quadratic equation.

In the exercise, since \( 4 \) and \( 5 \) are the solutions, the equation becomes \( a(x - 4)(x - 5) = 0 \). When we expand this, we get \( (x - 4)(x - 5) = x^2 - 9x + 20 \). The original equation \( ax^2 + bx + c = 0 \) thus takes the form \( ax^2 - 9ax + 20a = 0 \). This method of factoring quadratics helps to simplify the process and transition into easier to manage expressions.

Quadratic Coefficients

Understanding the coefficients \( a \), \( b \), and \( c \) in a quadratic equation \( ax^2 + bx + c = 0 \) is key to solving it. These coefficients are crucial as they dictate the shape and position of the quadratic equation's graph.

In our exercise, once we've expanded \( (x - 4)(x - 5) \) to get \( x^2 - 9x + 20 \), we want to match it back to the general quadratic form. By comparing \( ax^2 - 9ax + 20a = 0 \) with \( ax^2 + bx + c = 0 \), we can directly identify the coefficients: \( a = 1 \), \( b = -9 \), and \( c = 20 \).

These values tell us that the quadratic equation is symmetric around its vertex, and crosses the x-axis at the solutions \( x = 4 \) and \( x = 5 \). Mastery of these coefficients allows students to recognize and solve various forms of quadratic equations efficiently.