Problem 125 - Elementary and Intermediate Algebra

Question

Explain the error. Then find the correct result. a. Add: \(\quad \sqrt{-16}+\sqrt{-9}=\sqrt{-25}\) b. Multiply: \(\quad \sqrt{-2} V-3=\sqrt{-2(-3)}=\sqrt{6}\)

Step-by-Step Solution

Answer

a. Correct result: \(7i\); b. Correct result: \(-3i\sqrt{2}\).

Step 1: Identify the Error in Expression (a)

The expression given is to add two square roots of negative numbers: \( \sqrt{-16} \) and \( \sqrt{-9} \). Simplifying each as an imaginary number: \( \sqrt{-16} = 4i \) and \( \sqrt{-9} = 3i \). The mistake is adding these under the same square root: \( \sqrt{-25} \). The correct approach is to add the imaginary parts together: \( 4i + 3i = 7i \).

Step 2: Solve the Corrected Addition Expression (a)

Given \( \sqrt{-16} = 4i \) and \( \sqrt{-9} = 3i \), correctly sum the imaginary numbers: \( 4i + 3i = 7i \). Hence, the correct result is \( 7i \), not \( \sqrt{-25} \).

Step 3: Identify the Error in Expression (b)

The expression involves multiplying with \( \sqrt{-2} \). In the problem, the math is performed as \( \sqrt{-2}(-3) \), which incorrectly simplifies to \( \sqrt{6} \). Instead, the correct process involves distributing the \( \sqrt{-2} \), indicating a multiplication problem not directly under a square root.

Step 4: Solve the Correct Multiplication Expression (b)

To find the result of \( \sqrt{-2} \times (-3) \), first simplify \( \sqrt{-2} \) to \( i\sqrt{2} \). Then multiply: \( i\sqrt{2} \times (-3) = -3i\sqrt{2} \). Therefore, the correct result is \( -3i\sqrt{2} \), not \( \sqrt{6} \).

Key Concepts

Imaginary Numbers

Understanding imaginary numbers is critical when you work with the square roots of negative numbers. A key concept to grasp is that the square root of a negative number cannot be expressed as a real number because no real number squared gives a negative product.
Instead, we introduce imaginary numbers, which provide a solution. Here's how it works:

• The basic imaginary unit is denoted by \( i \), where \( i \) is defined as \( \sqrt{-1} \).
• Using this definition, the square root of any negative number can be expressed in terms of \( i \).
• For example, \( \sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i \).

When adding imaginary numbers, like in the exercise given, simply add the coefficients of \( i \). If you see \( 4i + 3i \) in a problem, your result will be \( 7i \). Each imaginary number's behavior aligns with the rules of algebraic expressions, just like other numbers, except they involve this unique unit \( i \).
This helps to solve equations that previously seemed impossible under real numbers alone.

Square Roots

When dealing with square roots, especially of negative numbers, it's crucial to understand how to switch your approach to accommodate imaginary numbers. The square root operation is essentially asking "what number, when squared, gives this product?"
To manage negative numbers under square roots properly:

• Recognize that square roots of negative numbers are undefined in the real number system but can be managed using \( i \).
• For example, \( \sqrt{-9} \) becomes \( 3i \).
• Remember that incorrect operations can lead to errors, such as when mistakenly simplifying \( \sqrt{-16} + \sqrt{-9} \) as \( \sqrt{-25} \).

Square root operations must be done carefully. Each radical simplification needs to convert negatives before joining them.
This process explains why adding under a single square root, as in the case of the problem faced, is incorrect. Imaginary numbers with square roots require their separate terms.

Algebraic Expressions

Algebraic expressions that involve imaginary numbers can be treated much like any other algebraic terms, but they come with rules unique to their nature.
When it comes to multiplying, like in the original exercise, there is a need to distribute terms just as you do in algebra, but recognize the presence of \( i \). Consider this:

• For multiplication involving \( \sqrt{-2} \), rewrite it as \( i\sqrt{2} \) first.
• Then, distribute normally: \( i\sqrt{2} \times (-3) = -3i\sqrt{2} \).
• This approach prevents errors like wrongly simplifying to \( \sqrt{6} \).

Algebraic expressions, thus, are less about the mystery of \( i \), and more about following procedure.
The essence lies in treating \( i\sqrt{n} \) as a variable form similar to \( a\sqrt{n} \) in real number expressions. Building these skills ensures success in manipulating expressions filled with both real and imaginary components.