Problem 98 - 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.) $$i,-i$$

Step-by-Step Solution

Answer

The values of \( a, b, \) and \( c \) are \( a = 1 \), \( b = 0 \), and \( c = 1 \).

Step 1: Understand the given solutions

The given solutions for the quadratic equation are complex numbers: \(i\) and \(-i\).

Step 2: Set up the quadratic equation

Use the zero-factor property in reverse. Since \(i\) and \(-i\) are solutions, the factors of the quadratic equation are \((x - i)\) and \((x + i)\).

Step 3: Form the quadratic equation

Multiply the factors to form the quadratic equation: \[(x - i)(x + i) = 0\]Apply the difference of squares formula: \[ (x - i)(x + i) = x^2 - (i)^2 = x^2 - (-1) = x^2 + 1 \].So the quadratic equation is: \[ x^2 + 1 = 0 \].

Step 4: Identify the coefficients

Compare \(x^2 + 1 = 0\) with the general form \(ax^2 + bx + c = 0\).This gives us: \[ a = 1, \ b = 0, \ c = 1 \].

Key Concepts

Complex Solutions

In quadratic equations, solutions can sometimes be complex numbers. Complex numbers include a real part and an imaginary part, represented by the symbol 'i', which stands for \( \sqrt{-1} \). For example, in the given exercise, the solutions to the quadratic equation were \(i\) and \(-i\).
Complex solutions often appear in conjugate pairs when dealing with real coefficients. A conjugate pair consists of two complex numbers that have the same real part but opposite imaginary parts. In our exercise, the solutions \(i\) and \(-i\) are a classic example of a conjugate pair. Understanding this concept is crucial as it often simplifies the process of forming quadratic equations.
To recap:

• Complex numbers have a real part and an imaginary part.
• Conjugate pairs have the same real part but opposite imaginary parts.

Zero-Factor Property

The zero-factor property is a key concept when solving quadratic equations. This property states that if the product of two factors is zero, then at least one of the factors must be zero.
In mathematical terms, if \(ab = 0\), then \(a = 0\) or \(b = 0\). This property is extremely useful because if you can express a quadratic equation as a product of factors, determining the roots becomes straightforward.
In the given exercise, we used the zero-factor property in reverse to understand that the solutions \(i\) and \(-i\) led us to the factors \((x - i)\) and \((x + i)\). Multiplying these factors allowed us to re-form the quadratic equation:
\( (x - i)(x + i) = 0 \).
Key points:

• The zero-factor property states that if \(ab = 0\), then \(a = 0\) or \(b = 0\).
• It simplifies solving quadratic equations by breaking them into factors.

Difference of Squares

The difference of squares formula is another essential tool in algebra and solving quadratic equations. It states that \((a - b)(a + b) = a^2 - b^2\). This formula allows us to simplify the multiplication of two binomials into a difference of squares.
In the given exercise, the factors were \((x - i)\) and \((x + i)\). When we applied the difference of squares formula, we obtained:
\((x - i)(x + i) = x^2 - i^2\).
Since \(i^2 = -1\), this becomes:
\( x^2 - (-1) = x^2 + 1 \).
Thus, the quadratic equation formed was \(x^2 + 1 = 0\). Recognizing and using the difference of squares formula can often simplify complex expressions, making it a vital technique in algebra.
To highlight:

• The difference of squares formula converts \((a - b)(a + b)\) into \(a^2 - b^2\).
• It is particularly useful for multiplying binomials where one term is the opposite of the other.