Problem 96 - 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.) $$-3,2$$

Step-by-Step Solution

Answer

a = 1, b = 1, c = -6

Step 1: Understand the given solutions

The quadratic equation has solutions -3 and 2. This means that when we factor the quadratic equation, it should look like a(x + 3)(x - 2) = 0.

Step 2: Use the zero-factor property in reverse

Since the solutions are -3 and 2, the factors of the quadratic equation are (x + 3) and (x - 2). Thus, the quadratic equation in its factored form is:a(x + 3)(x - 2) = 0.

Step 3: Expand the factors

Expand a(x + 3)(x - 2) to form a standard quadratic equation. Use the distributive property to get:a[(x)(x) + (x)(-2) + (3)(x) + (3)(-2)] = 0a(x^2 - 2x + 3x - 6) = 0aqx^2 + ax - 6a = 0.

Step 4: Identify the coefficients a, b, and c

From the expanded form, compare with the standard form of a quadratic equation ax^2 + bx + c = 0:a = ab = ac = -6a

Step 5: Choose convenient value for 'a'

For simplicity, let a = 1. Then,a = 1b = 1c = -6.

Key Concepts

Zero-Factor Property

The zero-factor property is a powerful tool in solving quadratic equations. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. In mathematical terms, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). This property is especially helpful in factoring quadratics because it allows us to break down complex equations into simpler ones.

For instance, given the quadratic equation with solutions -3 and 2, we know that \( (x + 3) \) and \( (x - 2) \) must be the factors. Therefore, \( (x + 3)(x - 2) = 0 \). By applying the zero-factor property, we can set each factor equal to zero:

• \( x + 3 = 0 \)
  
• \( x - 2 = 0 \)
  

Solving these, we find that \( x = -3 \) and \( x = 2 \). This reverse approach is useful when working with given solutions to find the original quadratic equation.

Factoring Quadratics

Factoring quadratics involves breaking down a quadratic equation into a product of two binomial expressions. This method is handy when you have solutions to the equation. For example, if the solutions are given as -3 and 2, these can be rewritten as factors of the quadratic.

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). To factor this, we need to reverse-process the solutions:

• \( (x + 3) \cdot (x - 2) \) represents the factored form based on solutions -3 and 2.
  
• For any quadratic equation with given solutions \( p \) and \( q \), the equation can be factored as \( (x - p)(x - q) = 0 \).
  

Through this process, we translate the solutions into their corresponding factors. Next, we can expand these factors to return to the standard quadratic form.

Solving Quadratics

Solving quadratic equations can be approached through various methods such as factoring, completing the square, or using the quadratic formula. In this exercise, we focus on solving by factoring.

With given solutions -3 and 2, we start by forming the factors: \( (x + 3) \) and \( (x - 2) \). We set up our factored equation:

• \( a(x + 3)(x - 2) = 0 \)
  

Expanding this, we get:

• \( a(x^2 - 2x + 3x - 6) = 0 \), which simplifies to \( a(x^2 + x - 6) = 0 \)
  

Comparing this with the standard form \( ax^2 + bx + c = 0 \), we identify:

• \( a = a \)
  
• \( b = a \)
  
• \( c = -6a \)
  

Setting \( a = 1 \) for simplicity, we find the specific coefficients as: \( a = 1 \), \( b = 1 \), and \( c = -6 \). Therefore, our original quadratic equation is \( x^2 + x - 6 = 0 \), aligning perfectly with the given solutions.