Problem 44 - A Graphical Approach to Precalculus with Limits

Question

Multiply or divide as indicated. Simplify each answer. $$\frac{\sqrt{-12} \cdot \sqrt{-6}}{\sqrt{8}}$$

Step-by-Step Solution

Answer

The simplified answer is -3.

Step 1: Recognize Imaginary Numbers

The expression contains square roots of negative numbers: \(\sqrt{-12}\) and \(\sqrt{-6}\). We need to rewrite these terms using \(i\), which is the imaginary unit defined as \(i = \sqrt{-1}\). Thus, \(\sqrt{-12} = \sqrt{-1} \cdot \sqrt{12} = i\sqrt{12}\) and \(\sqrt{-6} = \sqrt{-1} \cdot \sqrt{6} = i\sqrt{6}\).

Step 2: Multiply Imaginary Terms

Multiply the imaginary terms in the numerator: \(i\sqrt{12} \times i\sqrt{6} = i^2\sqrt{12} \cdot \sqrt{6}\). Since \(i^2 = -1\), the expression becomes \(-\sqrt{12 \times 6}\).

Step 3: Simplify the Radical Multiplication

Calculate \(12 \times 6 = 72\) and find \(-\sqrt{72}\). Now simplify \(\sqrt{72}\) by expressing 72 as a product of its factors: \(72 = 36 \times 2 = 6^2 \times 2\). Therefore, \(\sqrt{72} = \sqrt{6^2 \times 2} = 6\sqrt{2}\). This simplifies to \(-6\sqrt{2}\).

Step 4: Divide by \(\sqrt{8}\)

Now, divide \(-6\sqrt{2}\) by \(\sqrt{8}\), which is expressed as \(\sqrt{4 \times 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}\). Hence, \(-\frac{6\sqrt{2}}{2\sqrt{2}} = -\frac{6}{2} = -3\).

Step 5: Simplify Final Answer

The expression simplifies to \(-3\) as the \(\sqrt{2}\) terms cancel out in both the numerator and the denominator.

Key Concepts

Imaginary Numbers

Sometimes, we encounter the square root of negative numbers, which regular math doesn't allow us to simplify using real numbers. But that's where imaginary numbers come in handy. The symbol \( i \) is used to represent the square root of \(-1\), so whenever we have a square root of a negative number, we can express it using \( i \). For example, \( \sqrt{-12} \) becomes \( i \sqrt{12} \) because \(-12\) can be broken down into \( -1 \times 12 \). Thus, the \( \sqrt{-1} \) part becomes \( i \), and we are left with \( \sqrt{12} \). This conversion helps us work with negative square roots in a way that's consistent with the rules of arithmetic and algebra.

Simplifying Radicals

Radicals can sometimes look complicated, but simplifying them can make calculations much easier. To simplify a square root, you look for perfect square factors of the number inside the radical. For instance, with \( \sqrt{72} \), we know that \( 72 = 36 \times 2 \) and \( 36 \) is a perfect square. So we can rewrite \( \sqrt{72} \) as \( \sqrt{36} \cdot \sqrt{2} = 6 \sqrt{2} \). This process transforms the number inside the square root into simpler terms that are easier to work with in multiplication or division.

Multiplication and Division of Radicals

When multiplying radicals, you can multiply the numbers inside the radicals together. For instance, \( \sqrt{12} \times \sqrt{6} = \sqrt{12 \times 6} = \sqrt{72} \). Once you have the product, you can then simplify it as we did before. Division follows a similar principle. You can divide the numbers under the radicals:

• If you have \( \frac{\sqrt{a}}{\sqrt{b}} \), it simplifies to \( \sqrt{\frac{a}{b}} \) as long as \( b \) is not zero.
• For example, \( \frac{6\sqrt{2}}{\sqrt{8}} = \frac{6}{\sqrt{4 \times 2}} = \frac{6}{2\sqrt{2}} = 3 \).

Remember, rationalizing the denominator, which means getting rid of the radical in the denominator, might also be a step in simplifying fractions with radicals.

Square Roots of Negative Numbers

It's important to understand why square roots of negative numbers involve imaginary numbers. By definition, a square is a result of multiplying a number by itself. A negative square root cannot have a real number as its result because multiplying the same positive or negative number by itself always yields a positive product.

• A negative number's square root enters the realm of imaginary numbers. That's why we say \( \sqrt{-1} = i \).
• This approach simplifies complex calculations, like multiplying or dividing these non-real numbers, by allowing us to handle them using the usual algebraic rules.

Once you acknowledge \( i \), handling negative roots becomes manageable, making complex numbers integral to advanced mathematics.