Problem 27 - Algebra 2 and Trigonometry

Question

In \(19-34,\) write each sum or difference in terms of \(i\) $$ \sqrt{4}+\sqrt{-32}+\sqrt{-8} $$

Step-by-Step Solution

Answer

The sum in terms of \( i \) is \( 2 + 6\sqrt{2}i \).

Step 1: Simplify the Real Square Root

The square root of 4 is a real number. Simplify it to get \( \sqrt{4} = 2 \).

Step 2: Simplify the Imaginary Part of \( \sqrt{-32} \)

To simplify the square root of a negative number, separate the negative sign using \( i \), where \( i = \sqrt{-1} \). First, note that \( \sqrt{-32} = \sqrt{32} \cdot i \). We can further simplify \( \sqrt{32} \) by rewriting it as \( \sqrt{16 \cdot 2} = 4\sqrt{2} \), so \( \sqrt{-32} = 4\sqrt{2}i \).

Step 3: Simplify the Imaginary Part of \( \sqrt{-8} \)

Similarly, simplify \( \sqrt{-8} \) by separating the negative sign using \( i \). We have \( \sqrt{-8} = \sqrt{8} \cdot i \). Simplify \( \sqrt{8} \) as \( \sqrt{4 \cdot 2} = 2\sqrt{2} \), so \( \sqrt{-8} = 2\sqrt{2}i \).

Step 4: Combine All Parts

Combine the simplified expressions: \( 2 + 4\sqrt{2}i + 2\sqrt{2}i \).

Step 5: Simplify the Imaginary Parts

Combine the imaginary parts: \( 4\sqrt{2}i + 2\sqrt{2}i = 6\sqrt{2}i \).

Step 6: Write the Final Expression

The sum of all parts is \( 2 + 6\sqrt{2}i \).

Key Concepts

Imaginary Unit

In mathematics, dealing with square roots of negative numbers introduces the concept of the imaginary unit, represented as \( i \). The imaginary unit is essential for expressing numbers that the conventional real number system cannot handle. The definition of \( i \) is given as \( i = \sqrt{-1} \). When you square \( i \), you return to a familiar real number: \( i^2 = -1 \).

This intriguing property of \( i \) helps us manage otherwise undefined operations such as the square root of a negative number. Without \( i \), calculating expressions like \( \sqrt{-32} \) becomes impossible in the realm of real numbers. By utilizing \( i \), we expand our numerical system to include complex numbers, enabling us to perform more complex mathematical operations. Thus, for any negative number \( -a \), we use the formula \( \sqrt{-a} = \sqrt{a} \cdot i \) to find its square root.

Square Roots of Negative Numbers

Square roots are a fundamental operation in mathematics, but they become more interesting when applied to negative numbers. Normally, in the real number system, square roots of negative numbers do not exist because no real number squared gives a negative result. However, once we introduce the imaginary unit \( i \), calculating these square roots becomes possible.

For instance, to find \( \sqrt{-32} \), you break it apart by separating the negative part, using \( i \) as follows:

• First, rewrite \( \sqrt{-32} \) as \( \sqrt{32} \cdot i \).
• Then simplify \( \sqrt{32} \) by factoring into smaller known squares; here, \( 32 = 16 \times 2 \).
• Thus, \( \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \).
• Combine to get \( \sqrt{-32} = 4\sqrt{2}i \).

This approach can apply to any negative number, transforming it into a solvable form with \( i \). The key takeaway is that we employ \( i \) to bypass the limitations of real numbers, providing a bridge to more comprehensive solutions.

Simplifying Radicals

Simplifying radicals involves expressing a square root in its simplest form. This process helps to make complex calculations more manageable. When dealing with expressions such as \( \sqrt{32} \) or \( \sqrt{8} \), simplifying them can seem daunting, but it's quite straightforward with a few steps.

To simplify a radical, first factor the number into its prime components or as a product of squares:

• For \( \sqrt{32} \), recognize that \( 32 = 16 \times 2 \).
• This gives \( \sqrt{32} = \sqrt{16 \cdot 2} \), and because \( \sqrt{16} = 4 \), it simplifies to \( 4\sqrt{2} \).
• For \( \sqrt{8} \), use \( 8 = 4 \times 2 \) leading to \( \sqrt{8} = 2\sqrt{2} \).

In each case, you're looking for perfect squares inside the radical to simplify. Breaking down complicated square roots into simpler terms speeds up calculations and provides cleaner, more understandable results. With practice, simplifying radicals becomes a straightforward skill, beneficial in many areas of mathematics.