Problem 57 - Precalculus Mathematics for Calculus

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 \(x = 7i\) and \(x = -7i\).

Step 1: Identify the Equation Form

The given equation is \(x^2 + 49 = 0\). This is a quadratic equation in the form \(ax^2 + bx + c = 0\) with \(a = 1\), \(b = 0\), and \(c = 49\).

Step 2: Rearrange the Equation

Move the constant term to the other side of the equation: \(x^2 = -49\). This shows that \(x^2\) must be equal to \(-49\), which is a negative number.

Step 3: Solve for x Using Imaginary Numbers

To solve \(x^2 = -49\), take the square root of both sides. Remember that the square root of a negative number introduces the imaginary unit \(i\), where \(i^2 = -1\). Therefore:\[ x = \pm \sqrt{-49} = \pm \sqrt{49} \cdot \sqrt{-1} = \pm 7i \]

Step 4: Express Solutions in a + bi Form

The solutions are \(x = 7i\) and \(x = -7i\). Expressing these in the standard form of a complex number \(a + bi\), we have:\(x_1 = 0 + 7i\) and \(x_2 = 0 - 7i\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that involve a variable raised to the second power. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our original equation, \( x^2 + 49 = 0 \), we can see that \( a = 1 \), \( b = 0 \), and \( c = 49 \).

Solving a quadratic equation typically involves finding the values of \( x \) that satisfy the equation. These solutions can be found using several methods, such as factoring, completing the square, or employing the quadratic formula:

- Factoring involves expressing the quadratic expression as a product of two binomials.
- Completing the square involves rewriting the quadratic in the form \((x + p)^2 = q\).
- The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides a powerful tool to solve any quadratic equation.

The nature of the solutions depends on the discriminant, which is \( b^2 - 4ac \). If it's positive, there are two distinct real roots; if zero, one real root; and if negative, two complex roots.

Imaginary Unit

The imaginary unit \( i \) is an essential concept in complex numbers. It is defined as the square root of \(-1\), which means \( i^2 = -1 \). Mathematics uses \( i \) to help solve equations that involve square roots of negative numbers.

When dealing with quadratic equations like \( x^2 = -49 \), the emergence of the imaginary unit allows us to take the square root of a negative number. In this specific problem, the calculation \( \sqrt{-49} \) becomes \( \sqrt{49} \times \sqrt{-1} = 7i \).

The imaginary unit plays a crucial role in extending the number system beyond real numbers to include complex numbers, enabling solutions to a wider range of mathematical problems.

Complex Number Form a+bi

Complex numbers are numbers that have both a real part and an imaginary part, and they are expressed in the form \( a + bi \), where \( a \) is the real part, and \( b \) is the imaginary part.

These numbers stem from the need to handle square roots of negative numbers, which are resolved with the imaginary unit \( i \). For example, the solutions \( 7i \) and \( -7i \) to the equation \( x^2 + 49 = 0 \) can be expressed in this form. Here, \( a = 0 \) for both solutions as there is no real part, and \( b = \pm 7 \).

So, expressed in the standard form, the solutions are \( 0 + 7i \) and \( 0 - 7i \), neatly fitting the \( a + bi \) model. Complex numbers not only solve certain algebraic equations, but are also vital in fields such as engineering and physics, where they are used to describe phenomena that oscillate or rotate, such as waves and signals.