Problem 32 - Elementary Algebra

Question

\(w^{2}=-3\)

Step-by-Step Solution

Answer

w = ±i√(3)

Step 1 - Understand the Equation

The given equation is a quadratic equation, but instead of the usual form of ax^2 + bx + c = 0, it is given as w^2 = -3.

Step 2 - Isolate the Variable

Since the equation is already in the form w^2 = -3, the variable w is already isolated.

Step 3 - Recognize the Nature of the Solution

Notice that -3 is a negative number. The square of a real number is always non-negative, so w must be a complex number.

Step 4 - Solve for w Using Imaginary Numbers

To solve for w, take the square root of both sides of the equation. Recall that the square root of a negative number involves the imaginary unit i, where i = √(-1).

Step 5 - Apply the Square Root

Taking the square root of both sides, we get: w = ±√(-3) = ±√(3) * i. So, w = ±i√(3).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that involve variables raised to the second power. They generally have the form: \[ax^2 + bx + c = 0\] where \(a\), \(b\), and \(c\) are constants. In the given exercise, the equation is presented as \(w^2 = -3\). This is a special case where there's no linear term \(bx\) or constant term \(c\). Quadratic equations can be solved using various methods like:

• Factoring
• Completing the square
• Using the quadratic formula \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
• Graphing

In this specific case, where you need to find the square root of a negative number, complex numbers are involved. This leads us to our next concept.

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are expressed in the form: \(z = a + bi\) where \(a\) is the real part and \(bi\) is the imaginary part. Here, \(i\) stands for the imaginary unit, which is defined as \(i = \sqrt{-1}\). In the context of our example, solving for \(w\) in the equation \(w^2 = -3\) requires understanding that real numbers alone won't suffice, because the square of a real number can't be negative. Thus, we use complex numbers to provide a solution, and recognize that: \(\sqrt{-3} = \sqrt{3}i\). This gives us two solutions: \(w = +\sqrt{3}i\) and \(w = -\sqrt{3}i\). This is because taking the square root of both sides of the equation results in two values, one positive and one negative.

Square Root

The square root of a number \(x\) is a value \(y\) such that when \(y\) is multiplied by itself, it equals \(x\): \(y^2 = x\). For non-negative numbers, the square root is straightforward. For instance, \(\sqrt{9} = 3\), because \(3 \cdot 3 = 9\). But how do we handle negative numbers? When the square root of a negative number is required, such as \(\sqrt{-3}\), we introduce the imaginary unit \(i\), defined as \(i = \sqrt{-1}\). Thus: \(\sqrt{-3} = \sqrt{3i^2} = \sqrt{3}i\).So, taking the square root of both sides in our problem \(w^2 = -3\), we get: \(w = \pm i \sqrt{3}\) which encapsulates the idea that there are usually two solutions to a quadratic equation, even when it involves complex numbers.