Problem 23 - Algebra 2 and Trigonometry

Question

In \(19-34,\) write each sum or difference in terms of \(i\) $$ \sqrt{-12}+\sqrt{-27}-\sqrt{-75} $$

Step-by-Step Solution

Answer

The expression simplifies to 0.

Step 1: Understand the Problem

We need to simplify the expression \( \sqrt{-12} + \sqrt{-27} - \sqrt{-75} \) in terms of the imaginary unit \( i \), where \( i = \sqrt{-1} \). This involves expressing square roots of negative numbers using \( i \).

Step 2: Simplify Each Square Root

Rewrite each square root of a negative number in terms of \( i \):\[ \sqrt{-12} = \sqrt{12} \cdot \sqrt{-1} = \sqrt{12}i \] Similarly, \[ \sqrt{-27} = \sqrt{27} \cdot \sqrt{-1} = \sqrt{27}i \] and \[ \sqrt{-75} = \sqrt{75} \cdot \sqrt{-1} = \sqrt{75}i \]

Step 3: Simplify the Radicals

Now, simplify each radical:\[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \] So, \( \sqrt{12}i = 2\sqrt{3}i \).Similarly,\[ \sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3} \] So, \( \sqrt{27}i = 3\sqrt{3}i \).And, \[ \sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3} \] So, \( \sqrt{75}i = 5\sqrt{3}i \).

Step 4: Add and Subtract the Terms

Combine the simplified terms:\[ 2\sqrt{3}i + 3\sqrt{3}i - 5\sqrt{3}i \] Combine the coefficients:\[ (2 + 3 - 5)\sqrt{3}i = 0\sqrt{3}i = 0 \]

Step 5: Conclusion

The expression simplifies completely to \( 0 \), as the \( i \) terms cancel each other out.

Key Concepts

Complex Numbers

Complex numbers are a fascinating extension of the real numbers. They consist of two parts: a real part and an imaginary part. The real part is just like the numbers you are already familiar with, such as 2, -5, or 0. Meanwhile, the imaginary part involves the imaginary unit, denoted as \( i \), which is defined as the square root of \( -1 \).
For example, a complex number might look like \( 3 + 4i \), where 3 is the real part and \( 4i \) is the imaginary part.

• The imaginary part's unique quality is that when squared, it equals -1: \( i^2 = -1 \).
• Understanding complex numbers is useful in solving equations that don't have real number solutions.

When handling complex operations, remember:

• Add or subtract the real parts separately from the imaginary parts.
• When multiplying, use \( i^2 = -1 \) to simplify the result.

Complex numbers provide new ways to solve problems, particularly when completing operations that involve square roots of negative numbers.

Square Roots of Negative Numbers

Square roots of negative numbers can be tricky. In real numbers, taking the square root of a negative is impossible. However, with imaginary numbers, we can handle this easily using the concept of \( i \).
Let's consider what it means when you see \( \sqrt{-12} \). This is asking for a number that when squared gives -12. This is where the imaginary unit \( i \) comes into play:

• First, rewrite it as \( \sqrt{12} \times \sqrt{-1} \).
• Since \( \sqrt{-1} = i \), the expression becomes \( \sqrt{12}i \).

This method allows us to express any square root of a negative number in terms of \( i \).
By doing so, we can work with them almost as easily as we work with real numbers. Just remember to handle \( i \) as representing \( \sqrt{-1} \) whenever it appears in your calculations.

Simplifying Radicals

Simplifying radicals involves breaking down the number inside the radical into its prime factors and finding perfect square factors. Once you identify the perfect square, you can remove it from the radical.
For example, to simplify \( \sqrt{12} \):

• Factor 12 into \( 4 \times 3 \).
• The square root of 4 is 2, so \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \).

This technique applies similarly to other radicals, such as \( \sqrt{27} \) and \( \sqrt{75} \), where you look for the largest perfect square factor:

• \( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \)
• \( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \)

Simplifying radicals not only makes computations easier, but also helps in combining similar terms. This comes in handy when adding or subtracting complex expressions, like in the given mathematical expression.