Problem 89 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$(x-2)^{2}=-9$$

Step-by-Step Solution

Answer

x = 2 + 3i, x = 2 - 3i

Step 1: Understand the Equation

The given equation is \( (x-2)^2 = -9 \). Notice that it is a quadratic equation set equal to a negative number.

Step 2: Isolate the Square

Rewrite the equation as \( (x-2)^2 = -9 \). To find the solutions, we need to take the square root of both sides. First, note that the square root of a negative number involves imaginary numbers.

Step 3: Apply Square Roots

Taking the square root of both sides, we get \( \sqrt{(x-2)^2} = \sqrt{-9} \). This simplifies to \( x-2 = \pm \sqrt{-9} \).

Step 4: Simplify the Square Root of a Negative Number

Recognize that \( \sqrt{-9} = 3i \), where \ i \ is the imaginary unit. Therefore, \[ x - 2 = \pm 3i \].

Step 5: Solve for x

Isolate \x\ by adding 2 to both sides of the equations \x - 2 = 3i\ and \x - 2 = -3i\. This gives: \[ x = 2 + 3i \] and \[ x = 2 - 3i \].

Key Concepts

quadratic equations

Quadratic equations are polynomial equations of degree 2, which means they have the form \(ax^2 + bx + c = 0\). They are characterized by the highest exponent of the variable being 2. Quadratic equations can have different types of solutions depending on the discriminant (\(b^2 - 4ac\)).

Here are the possible cases:

• If the discriminant is positive, there are two distinct real solutions.
• If the discriminant is zero, there is one real, repeated solution.
• If the discriminant is negative, there are two complex (or imaginary) solutions.

The most common methods to solve these equations are:

• Factoring
• Completing the square
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

A good understanding of quadratic equations is essential for solving a wide variety of problems in mathematics.

imaginary solutions

Imaginary solutions occur when the discriminant of a quadratic equation is negative. The discriminant (\(b^2 - 4ac\)) being negative means the equation does not intersect the x-axis and has no real solutions.

Instead, we deal with complex numbers involving the imaginary unit \(i\), where \(i^2 = -1\). For example, consider the equation \((x-2)^2 = -9\):

• First, rewrite it as \((x-2)^2 = -9\).
• Take the square root of both sides: \(\sqrt{(x-2)^2} = \sqrt{-9}\)
• This simplifies to \(x-2 = \pm \sqrt{-9}\)
• Recognize that \(\sqrt{-9} = 3i\)
• This gives two solutions: \(x = 2 + 3i\) and \(x = 2 - 3i\)

Thus, the imaginary solutions to our example come out to be complex numbers. Understanding how to handle the imaginary unit \(i\) and work with complex solutions is key in advanced mathematics.

square roots

Square roots are mathematical operations that reverse the squaring of a number. If \(y = x^2\), then the square root of \(y\) is \(\sqrt{y} = x\) or \(\pm x\). When dealing with positive numbers, the square root operation is straightforward. For instance: \(\sqrt{9} = 3\) because \(3^2 = 9\).

However, taking square roots of negative numbers involves imaginary numbers. Since no real number squared will result in a negative value, we introduce the imaginary unit \(i\). Here, \(i\) is defined as \(\sqrt{-1}\), leading us to \(i^2 = -1\).

• Example: \(\sqrt{-9}\) can be written as \(\sqrt{-1 \cdot 9}\).
• This separates to: \(\sqrt{-1} \cdot \sqrt{9} = i \cdot 3\).
• Therefore, \(\sqrt{-9} = 3i\).

To sum up, understanding square roots and their properties, especially involving negative numbers, is crucial for solving problems with imaginary and real components.