Problem 68 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$w^{2}+9=0$$

Step-by-Step Solution

Answer

\(w = 3i\) or \(w = -3i\)

Step 1 - Rewrite the Equation

Start by rewriting the equation in a more convenient form. The given equation is:\[w^2 + 9 = 0\]

Step 2 - Isolate the Quadratic Term

Subtract 9 from both sides to isolate the quadratic term:\[w^2 = -9\]

Step 3 - Take the Square Root of Both Sides

Take the square root of both sides to solve for \(w\). Remember to include the positive and negative roots:\[w = \pm \sqrt{-9}\]

Step 4 - Simplify the Square Root

Simplify the square root of \(-9\). Recall that \(\sqrt{-1} = i\), where \(i\) is the imaginary unit:\[w = \pm \sqrt{9} \cdot \sqrt{-1} = \pm 3i\]

Step 5 - State the Solutions

The solutions to the equation are:\[w = 3i \text{ or } w = -3i\]

Key Concepts

Quadratic Equation

A quadratic equation is any equation that can be expressed in the form \[\begin{equation}ax^2 + bx + c = 0\end{equation}\]where \(a\), \(b\), and \(c\) are constants.In our original exercise, the equation given is \[\begin{equation} w^2 + 9 = 0 \end{equation}\]. This is a simple quadratic equation since it can be rewritten in the standard form.

Quadratic equations can have:

• two real solutions
• one real solution
• two imaginary solutions

We encounter the latter in this exercise. This categorization depends on the discriminant \[\begin{equation}b^2 - 4ac\end{equation}\].

If the discriminant is negative, the equation has imaginary solutions, as in our case where it simplifies to \[\begin{equation}w^2 = -9\end{equation}\].

Complex Numbers

Complex numbers include a real part and an imaginary part. They are expressed as \[\begin{equation}a + bi\end{equation}\], where \(a\) is the real part and \(bi\) is the imaginary part. In our solution for \[\begin{equation}w^2 + 9 = 0\end{equation}\], we find that \(w\) can be \(±3i\).

This means our solutions are purely imaginary numbers. The imaginary unit, represented by \(i\), is defined as \[\begin{equation}i^2 = -1\end{equation}\]. This is crucial when dealing with square roots of negative numbers as it helps us simplify expressions involving imaginary solutions.

Square Root

Taking the square root of a number means finding a value that, when multiplied by itself, gives the original number. Normally, the square root of a positive number, like \(9\), will yield real solutions: \[\begin{equation} ±3 \end{equation}\] However, when dealing with negative numbers under the square root, you get imaginary solutions.

In this exercise, the expression \[\begin{equation} \(±\text{sqrt(-9)}\) \end{equation}\] simplifies as follows:\[\begin{equation} \(± \text{sqrt(9) * sqrt(-1)}\) = \(± 3i\) \end{equation}\].

This transformation is made possible by \[\begin{equation}i, which is $$\text{sqrt(-1)}. \end{equation}\]Understanding how to manipulate square roots when they involve negative numbers is key to solving equations involving complex numbers.