Problem 56 - Precalculus Mathematics for Calculus

Question

Evaluate the radical expression and express the result in the form \(a+b i\) $$\frac{\sqrt{-7} \sqrt{-49}}{\sqrt{28}}$$

Step-by-Step Solution

Answer

The expression evaluates to \(-\frac{7}{2} + 0i\).

Step 1: Express the square roots of negative numbers using the imaginary unit

First, rewrite the square roots of negative values in terms of the imaginary unit, where \(i = \sqrt{-1}\). So, \(\sqrt{-7} = \sqrt{7} \cdot i\) and \(\sqrt{-49} = \sqrt{49} \cdot i\). This gives us \(\sqrt{49} = 7\), thus, \(\sqrt{-49} = 7i\).

Step 2: Simplify the expression by substituting the imaginary units

Substitute the results from Step 1 into the original expression: \(\frac{(\sqrt{7} \cdot i)(7i)}{\sqrt{28}}\).

Step 3: Multiply and simplify the terms in the numerator

Multiply the terms in the numerator: \((\sqrt{7} \cdot 7)(i \cdot i) = 7\sqrt{7} \cdot (-1) = -7\sqrt{7}\).

Step 4: Simplify the denominator

Simplify \(\sqrt{28}\) by rewriting it in terms of simpler square roots: \(\sqrt{28} = \sqrt{4 \cdot 7} = \sqrt{4} \cdot \sqrt{7} = 2\sqrt{7}\).

Step 5: Simplify the entire expression

Divide the simplified numerator by the simplified denominator: \(- \frac{7\sqrt{7}}{2\sqrt{7}}\). The \(\sqrt{7}\) terms cancel out. So we have \(-\frac{7}{2}\).

Step 6: Express the final result in standard form

Since the imaginary part of the number is 0, express the result as \(-\frac{7}{2} + 0i\).

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit is a fundamental concept used to extend the real number system.
It is denoted by the symbol \(i\) and is defined as the square root of \(-1\). This means that \(i^2 = -1\).
The imaginary unit allows us to handle the square roots of negative numbers, which is otherwise not possible within the real number system.
With the imaginary unit, we can express any complex number in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

• \(a\) is called the real part.
• \(b\) is called the imaginary part.

The concept of the imaginary unit is not just mathematical curiosity; it is essential in fields like engineering, physics, and signal processing. It enables solving equations like \(x^2 + 1 = 0\) and allows us to work with electrical currents and waves, which often have oscillatory nature.

Radical Expressions

Radical expressions involve roots, such as square roots, cube roots, and so on.
A radical expression typically has a radical sign \(\sqrt{}\), and the number or expression inside this symbol is the radicand. Understanding radicals is key when dealing with expressions involving the square roots of negative numbers.
For example, the expression \(\sqrt{-7}\) cannot be simplified within the real numbers because a square root normally refers to non-negative solutions.

• To manage these, we use the imaginary unit: \(\sqrt{-7} = \sqrt{7} \cdot i\).
• This conversion is crucial because it allows us to perform operations with these terms.

Often, radical expressions also require factorization to simplify.
Understanding how to break down and combine these parts is vital when simplifying complex numbers or expressions. Knowing that \(i = \sqrt{-1}\), any square root of a negative number can be rewritten using \(i\), making it more manageable.

Simplifying Expressions

Expression simplification is an important skill that involves reducing an expression to its simplest form.
Simplifying often means factoring, canceling, or combining like terms so that the expression becomes easier to understand and work with.
In the context of complex numbers, simplification can involve working through expressions with imaginary units and radicals.

• One approach involves simplifying the terms inside the expression, like breaking down square roots into their prime factors.
• We also use the property of the imaginary unit \(i^2 = -1\) to manage terms involving \(i\).

For example, in an expression like \(-\frac{7\sqrt{7}}{2\sqrt{7}}\), we can cancel the common term \(\sqrt{7}\) from the numerator and denominator, leading to a much simpler form \(-\frac{7}{2}\).
This simplification helps us express the result in standard form \(a + bi\), as was shown when the problem's solution was re-framed as \(-\frac{7}{2} + 0i\). Achieving this form not only provides clarity but also aids in further mathematical processes and applications.