Problem 58 - Algebra and Trigonometry

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

Step 1: Recognize the Equation

The given equation is \(9x^2 + 4 = 0\). Notice that this is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 9\), \(b = 0\), and \(c = 4\).

Step 2: Rearrange the Equation

We can rearrange the equation to isolate the \(x^2\) term. This gives us the equation \(9x^2 = -4\).

Step 3: Solve for \(x^2\)

Divide both sides of the equation by 9 to solve for \(x^2\):\[x^2 = -\frac{4}{9}\]This represents the square of \(x\) being equal to a negative number.

Step 4: Find \(x\) Using the Square Root

To find \(x\), we take the square root of both sides. Since the equation is equal to a negative number, the solutions will involve imaginary numbers:\[x = \pm \sqrt{-\frac{4}{9}} = \pm \frac{2i}{3}\]This is because \(\sqrt{-1} = i\), thus \(\sqrt{-\frac{4}{9}} = \frac{2i}{3}\).

Step 5: Express Solutions in \(a + bi\) Form

The solutions \(x = \frac{2i}{3}\) and \(x = -\frac{2i}{3}\) can be expressed in the form \(a + bi\) where \(a = 0\) and \(b = \frac{2}{3}\) for the first solution, and \(b = -\frac{2}{3}\) for the second solution.Thus, the solutions are \(x = 0 + \frac{2}{3}i\) and \(x = 0 - \frac{2}{3}i\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation of degree two, which typically take the form \( ax^2 + bx + c = 0 \). In this expression, \( a \), \( b \), and \( c \) are coefficients, and \( x \) represents the unknown variable we need to solve for.
Quadratic equations can have different types of solutions depending on the discriminant \( b^2 - 4ac \):

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If negative, the solutions are complex numbers involving imaginary components.

For the given equation \( 9x^2 + 4 = 0 \), there's no linear term (\( bx \)), and the discriminant yields a negative value \( (0 - 4 \, \times \, 9 \, \times \, 4) \), indicating that the solutions involve imaginary numbers.

Imaginary Numbers

Imaginary numbers are utilized when dealing with the square roots of negative numbers, offering a solution beyond the scope of real numbers.
The unit imaginary number is denoted as \( i \), where:\[ i = \sqrt{-1} \]
When solving quadratic equations like \( 9x^2 + 4 = 0 \), and upon rearranging to \( x^2 = -\frac{4}{9} \), we encounter the need for imaginary numbers. The use of \( i \) helps us express the solutions as:

• \( x = \pm \frac{2i}{3} \)

This expression shows that each solution is an imaginary number since they include the \( i \) component, indicating a solution on the imaginary axis when plotted on the complex plane.

Polar Form

Complex numbers, often expressed in the form \( a + bi \), can also be represented in polar form. This is particularly useful when visualizing or performing operations on these numbers. The polar form expresses a complex number using its magnitude \( r \) and angle \( \theta \) on the complex plane, given by \( z = r(\cos\theta + i\sin\theta) \). Alternatively, Euler’s formula can be applied: \( z = re^{i\theta} \).To convert \( x = \frac{2i}{3} \) to polar form, notice that:

• The magnitude \( r = \sqrt{(0)^2 + \left(\frac{2}{3}\right)^2} = \frac{2}{3} \)
• The angle \( \theta \) aligns with the positive imaginary axis, meaning \( \theta = \frac{\pi}{2} \)

Thus, the polar form for \( \frac{2i}{3} \) is \( \frac{2}{3}(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) \), or \( \frac{2}{3}e^{i\frac{\pi}{2}} \). Similarly, for \( -\frac{2i}{3} \), the angle \( \theta = \frac{-\pi}{2} \) would be used.