Problem 67 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$w^{2}+4=0$$

Step-by-Step Solution

Answer

The solutions are \( w = 2i \) and \( w = -2i \).

Step 1: Rewrite the Equation

Start by moving the constant term to the other side of the equation. The equation is given as: \[ w^2 + 4 = 0 \] Subtract 4 from both sides to get: \[ w^2 = -4 \]

Step 2: Take the Square Root of Both Sides

To solve for \( w \), take the square root of both sides of the equation: \[ w = \pm \sqrt{-4} \]

Step 3: Simplify the Square Root

Simplify the square root of -4. Recall that \( \sqrt{-1} = i \) (where \( i \) is the imaginary unit), thus: \[ w = \pm \sqrt{4} \cdot \sqrt{-1} = \pm 2i \]

Step 4: Write Final Solutions

The solutions to the equation are: \[ w = 2i \quad \text{and} \quad w = -2i \]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). In this form:

• \(ax^2\) is the quadratic term
• \(bx\) is the linear term
• \(c\) is the constant term

Quadratic equations can have real or complex solutions. Real solutions occur when the discriminant (\(b^2 - 4ac\)) is greater than or equal to zero. Complex solutions arise when the discriminant is less than zero, resulting in imaginary numbers.

In the given equation, \(w^2 + 4 = 0\), we rearranged it into a standard quadratic form: \(w^2 = -4\). This equation has no real solutions because the value under the square root is negative, leading us to work with imaginary numbers.

Complex Numbers

Complex numbers extend the concept of the one-dimensional number line. They form a two-dimensional complex plane. A complex number has a real part and an imaginary part and is generally written in the form \(a + bi\), where:

• \(a\) is the real part
• \(bi\) is the imaginary part

Here, \(i\) is the imaginary unit, defined as \(i = \sqrt{-1}\). Complex numbers are useful for solving equations that do not have real solutions. For example, the solution \(w = \pm 2i\) from our original equation arises because we need to find the square root of a negative number. As \(\sqrt{-4} = 2i\), we obtained two solutions: \(2i\) and \(-2i\).

Square Root

The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\). For example, \(\sqrt{9} = 3\) because \(3^2 = 9\). However, taking the square root of a negative number involves imaginary numbers. Specifically, \(\sqrt{-1} = i\). To handle the square root of other negative numbers, we use:

• \(\sqrt{-x} = \sqrt{x} \cdot i\)

In the solution, we saw that \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\). This simplification allowed us to find the solutions of the quadratic equation as \(\pm 2i\). Understanding the concept of square roots, both real and imaginary, is crucial for solving equations involving complex numbers.