Problem 66 - Contemporary Precalculus

Question

Simplify the expression without using a calculator. Your answer should not have any radicals in it. $$\sqrt{12}(\sqrt{3}-\sqrt{27})$$

Step-by-Step Solution

Answer

Answer: The simplified equivalent expression without radicals for the given expression is $$-12$$.

Step 1: Simplify Radicals inside Parentheses

We need to simplify $$\sqrt{27}$$. We can do this by thinking of its prime factorization: $$\sqrt{27} = \sqrt{3^3} = 3\sqrt{3}$$ Now substitute this back into the original expression: $$\sqrt{12}(\sqrt{3}-3\sqrt{3})$$

Step 2: Distribute $$\sqrt{12}$$ to Inside Terms

Now we distribute the $$\sqrt{12}$$ to both terms inside the parentheses: $$\sqrt{12} * \sqrt{3} - \sqrt{12} * 3\sqrt{3}$$

Step 3: Simplify Radicals and Combine Terms

First, let's simplify each radical separately: For $$\sqrt{12} * \sqrt{3}$$: $$\sqrt{12} * \sqrt{3} = \sqrt{36} = 6$$ For $$\sqrt{12} * 3\sqrt{3}$$: We can simplify $$\sqrt{12}$$ first, such that: $$\sqrt{12} = \sqrt{2^2 * 3} = 2\sqrt{3}$$ Substitute back into the second term: $$2\sqrt{3} * 3\sqrt{3} = 6(\sqrt{3})^2 = 6(3) = 18$$ Now the expression becomes: $$6 - 18$$

Step 4: Evaluate Final Expression

Evaluate the simplified expression: $$6 - 18 = -12$$ So, the simplest equivalent expression without radicals is: $$-12$$

Key Concepts

Simplify Square Roots

Simplifying square roots is one of the foundational skills in precalculus and algebra. It is the process of finding the simplest radical form of a given square root. For example, consider the square root of 12, \( \sqrt{12} \). To simplify it, we search for perfect square factors of 12. The number 12 can be expressed as \( 2^2 \) multiplied by 3, so \( \sqrt{12} = \sqrt{2^2 \cdot 3} = 2\sqrt{3} \).

Remember, whenever you encounter a square root, look for the largest square factor to simplify it. This makes the subsequent steps in the problem easier to manage and gets you to the simplest form faster. For instance, the number 27 can be broken down into \( 3^3 \) which simplifies to \($3\sqrt{3}\) by taking out a pair of threes outside the square root. In summary, always factorize the number under the square root and pull out perfect squares to simplify the radical.

Radical Expressions in Precalculus

Working with radical expressions is a common occurrence in precalculus. A radical expression is an expression that contains a square root, cube root, or higher roots. When you encounter a problem like \( \sqrt{12}(\sqrt{3}-\sqrt{27}) \), you're seeing a multiplication of radicals, which requires you to understand both how to simplify individual radicals and how to multiply them.

To tackle these expressions, start by simplifying each radical. If the radicands—the numbers inside the radical sign—are not perfect squares, reduce them by factoring out squares, much like you simplify fractions by canceling common factors. For the square root of 27, recognizing that 27 is \( 3^3 \) allows you to simplify it to \( 3\sqrt{3} \). By simplifying the radical components first, subsequent mathematical operations, like multiplication and addition, become significantly more straightforward.

Distributive Property with Radicals

The distributive property is a core principle in algebra that also applies when you’re working with radicals. It states that for any three numbers, a, b, and c, the expression \( a(b + c) = ab + ac \). This property also helps streamline the process of simplifying radical expressions.

In the context of your exercise, this means multiplying the \( \sqrt{12} \) with both terms within the parentheses, resulting in \( \sqrt{12} \cdot \sqrt{3} \) and \( \sqrt{12} \cdot 3\sqrt{3} \). Before you proceed, always simplify each of the radicals. Just like distributing numbers, the key is to distribute the radical to each term in the bracket and simplify where possible. In this scenario, we simplified \( \sqrt{12} \cdot 3\sqrt{3} \) by breaking down \( \sqrt{12} \) into \( 2\sqrt{3} \) before distributing. Finally, combine the terms, if possible, to conclude the simplification process. Understanding this property makes dealing with radical expressions less daunting and enables you to navigate through more complex algebra with confidence.