Problem 57 - Algebra and Trigonometry

Question

Find all solutions of the equation, and express them in the form \(a+b i\) $$ x^{2}+49=0 $$

Step-by-Step Solution

Answer

The solutions are \(0+7i\) and \(0-7i\).

Step 1: Rearrange the Equation

We start with the equation \( x^2 + 49 = 0 \). The first step is to isolate \( x^2 \) by subtracting 49 from both sides: \[ x^2 = -49 \] This sets up the equation to solve for \( x \).

Step 2: Solve for x Using the Square Root

The next step is to solve \( x^2 = -49 \) by taking the square root of both sides. Recall that the square root of a negative number involves the imaginary unit \( i \), where \( i = \sqrt{-1} \). Therefore, \[ x = \pm \sqrt{-49} = \pm 7i \] The solutions are imaginary numbers.

Step 3: Express Solutions in the Form of a + bi

The solutions \( x = 7i \) and \( x = -7i \) need to be expressed in the form \( a + bi \). Thus, we can write: \[ x = 0 + 7i \] and \[ x = 0 - 7i \] In both cases, \( a = 0 \) and \( b = 7 \) or \( b = -7 \).

Key Concepts

Quadratic Equations

Quadratic equations are an important type of polynomial equation that appear in the form \( ax^2 + bx + c = 0 \). In these equations, \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. Quadratic equations can take different forms:

• Standard Form: \( ax^2 + bx + c = 0 \)
• Factored Form: \( a(x - r)(x - s) = 0 \)
• Vertex Form: \( a(x - h)^2 + k = 0 \)

Solving a quadratic equation involves finding the values of \( x \) that make the equation true.
These values are called the roots of the equation. There are several methods to solve them:

• Factoring, when the equation can be transformed into the product of two binomials.
• Completing the square, which involves rearranging the equation to form a perfect square trinomial.
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Quadratic equations often have two solutions because the graph of a quadratic equation is a parabola.
These solutions can be real or complex numbers.

Imaginary Unit

The imaginary unit, denoted as \( i \), is a concept used in mathematics to extend the real number system. It is defined by the property \( i^2 = -1 \). This definition allows mathematicians to work with numbers that are not real, called imaginary numbers.

Imaginary numbers are used to solve equations that have no real solutions. They are crucial in fields such as engineering, physics, and computer science, where they simplify calculations involving oscillations, waves, and other complex systems.

• An imaginary number is typically expressed in the form \( bi \), where \( b \) is a real number.
• When added to a real number, it forms a complex number: \( a + bi \).

Using the imaginary unit, mathematicians can extend the solutions of quadratic equations. For instance, in the equation \( x^2 = -49 \), the solution involves \( \pm 7i \), demonstrating the use of the imaginary unit.

Square Root of Negative Numbers

The square root of a negative number is a concept that leads us to the realm of imaginary numbers. Normally, the square root of a negative number doesn't exist in the real number system. However, by using the imaginary unit \( i \), we can express these roots as imaginary numbers, making them part of the complex number system.

Here's how it works:

• To find the square root of \( -1 \), we define it as \( i \).
• Consequently, the square root solution extends to any negative number. For example, \( \sqrt{-49} = \sqrt{49} \cdot \sqrt{-1} = 7i \).

This rule helps in solving quadratic equations where the discriminant \( (b^2 - 4ac) \) is negative. Such solutions are not viable in the real number system, but the introduction of \( i \) allows them to be expressed as complex numbers, which include real and imaginary parts.
This extension is vital to understand the full spectrum of solutions in mathematical problems involving quadratics or other polynomial equations.