Problem 58 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$9 x^{2}+4=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{2}{3}i\) and \(x = -\frac{2}{3}i\).

Step 1: Isolate the Quadratic Term

The given equation is \(9x^2 + 4 = 0\). Begin by subtracting 4 from both sides to isolate the quadratic term: \(9x^2 = -4\).

Step 2: Divide by the Coefficient of \(x^2\)

Divide both sides of the equation by 9 to solve for \(x^2\): \(x^2 = \frac{-4}{9}\).

Step 3: Solve for x Using Square Roots

Take the square root of both sides of the equation.\[x = \pm \sqrt{\frac{-4}{9}}.\]Recognize that \(\sqrt{-1} = i\), the imaginary unit. Thus, the solutions are \(x = \pm \frac{2i}{3}\).

Step 4: Express in Complex Form

Finally, express each solution as a complex number in the form \(a+bi\). There are two solutions: \(x = \frac{2}{3}i\) and \(x = -\frac{2}{3}i\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation with a degree of 2, which means they usually take the form \( ax^2 + bx + c = 0 \). In this specific problem, we're dealing with a simpler form: \( 9x^2 + 4 = 0 \). Quadratic equations often represent parabolas when graphed on a coordinate plane. To solve them, we typically aim to isolate the \( x^2 \) term. Once isolated, we can apply methods like factoring, completing the square, or using the quadratic formula. However, in certain cases, like this exercise, it's clear that we end up with a negative number under the square root sign. This introduces the concept of complex numbers as potential solutions, showcasing an important branch of algebra.

Imaginary Unit

The imaginary unit, denoted as \( i \), is a fundamental concept in the study of complex numbers. It's defined by the equation \( i^2 = -1 \). This means that \( i \) is the number that when squared, equals \(-1\).In this exercise, when we took the square root of both sides of the equation \( x^2 = \frac{-4}{9} \), we encountered the need to calculate \( \sqrt{-1} \), which is where \( i \) comes into play. By using \( i \), we can still express solutions as numbers, even when dealing with the square root of a negative number.Key points about the imaginary unit:

• \( i^2 = -1 \)
• Imaginary numbers can be combined with real numbers to form complex numbers.
• \( i \) enables us to handle calculations that involve negative roots.

Complex Solutions

Complex solutions extend the idea of solving equations to include numbers that are not just real, but also imaginary. The typical form used to express complex numbers is \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part.In the given equation \( 9x^2 + 4 = 0 \), we derived solutions after isolating \( x^2 \) and recognizing the presence of an imaginary number. This led us to the solutions \( x = \frac{2}{3}i \) and \( x = -\frac{2}{3}i \). Both can be expressed as complex numbers:

• \( x = 0 + \frac{2}{3}i \)
• \( x = 0 - \frac{2}{3}i \)

These solutions highlight the combination of algebraic operations and the inclusion of imaginary units to solve quadratic equations that do not have real-number solutions. Complex solutions are particularly useful in fields like engineering and physics, where such calculations often occur.