Problem 20 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$x^{2}=-32$$

Step-by-Step Solution

Answer

There are no real solutions because a square of a real number cannot be negative.

Step 1: Analyze the Equation

Given the equation is \(x^2 = -32\). The equation represents a quadratic equation in the form \(x^2 = c\), where \(c = -32\).

Step 2: Evaluate for Real Solutions

Understand that \(x^2\) represents a square of a real number. The square of any real number is non-negative, meaning it cannot be negative. Since \(-32\) is less than zero, there are no real numbers \(x\) such that \(x^2 = -32\).

Step 3: Conclude No Real Solutions

From the analysis, conclude that since no real number squared gives a negative number, \(x^2 = -32\) has no real solutions.

Key Concepts

Complex Numbers

The equation \(x^2 = -32\) introduces the concept of complex numbers. Complex numbers expand the number system beyond real numbers, allowing for solutions to equations that would otherwise have no real solution. Complex numbers have the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The term \(i\) represents the imaginary unit, defined as \(i^2 = -1\).

• All real numbers are complex numbers with \(b = 0\).
• The imaginary part, \(bi\), is what helps us solve equations like \(x^2 = -32\).

To solve \(x^2 = -32\), we look for a number \(x\) that, when squared, equals \(-32\). Recognizing that \(i^2 = -1\), we rewrite \(-32\) as \(-1 \times 32\). So, we find the square root of \(-32\) using complex numbers: \(x = \pm \sqrt{-1 \times 32} = \pm i\sqrt{32}\). Simplifying further gives \(x = \pm 4i\sqrt{2}\).
This solution exemplifies how complex numbers enable us to solve equations that cannot be resolved within the real number system.

Real Numbers

Real numbers include all the numbers on the number line, extending infinitely in both positive and negative directions. They include integers, fractions, and irrationals, but importantly, they do not include the square roots of negative numbers. Such numbers require the introduction of complex numbers. In the exercise \(x^2 = -32\), we attempted to find a real number \(x\) that satisfies this equation.

• The square of any real number is always non-negative because the product of two positive numbers or two negative numbers is positive.
• This means that negative numbers like \(-32\) cannot be the square of a real number.

Thus, the equation \(x^2 = -32\) has no solutions within the scope of real numbers alone. This highlights the necessity of complex numbers when dealing with equations involving the square roots of negative numbers.

Imaginary Solutions

Imaginary solutions arise from equations that require us to take the square root of a negative number, something not possible within the set of real numbers alone. To handle such situations, we use the imaginary unit \(i\).\

• This is often seen in quadratic equations where the discriminant is negative.
• The term \(i\) allows us to "step outside" the real number system and find solutions to otherwise unsolvable equations.

For the quadratic equation \(x^2 = -32\), solving it yields imaginary solutions \(x = \pm 4i\sqrt{2}\). Here, the presence of \(i\) indicates that these solutions are not real numbers we encounter daily, but still valid within the mathematics encompassing complex numbers. Imaginary solutions showcase the power of the expanded number system by providing answers where real numbers fall short.