Problem 3 - Intermediate Algebra

Question

For Problems \(1-20\), use the distributive property to help simplify each of the following. For example, $$ \begin{aligned} 3 \sqrt{8}-\sqrt{32} &=3 \sqrt{4} \sqrt{2}-\sqrt{16} \sqrt{2} \\ &=3(2) \sqrt{2}-4 \sqrt{2} \\ &=6 \sqrt{2}-4 \sqrt{2} \\ &=(6-4) \sqrt{2}=2 \sqrt{2} \end{aligned} $$ $$ 7 \sqrt{12}+10 \sqrt{48} $$

Step-by-Step Solution

Answer

The simplified expression is \( 54 \sqrt{3} \).

Step 1: Express the Radicals with Perfect Squares

First, we decompose each square root expression into a product of a perfect square and a remaining number. For \( 7\sqrt{12} \), we write it as \( 7\sqrt{4 \times 3} = 7 \sqrt{4} \times \sqrt{3} \). For \( 10\sqrt{48} \), we have \( 10\sqrt{16 \times 3} = 10 \sqrt{16} \times \sqrt{3} \).

Step 2: Simplify the Radical Components

Next, we simplify the perfect square components. Since \( \sqrt{4} = 2 \) and \( \sqrt{16} = 4 \), the expressions simplify to \( 7 \times 2 \sqrt{3} = 14 \sqrt{3} \) and \( 10 \times 4 \sqrt{3} = 40 \sqrt{3} \).

Step 3: Use the Distributive Property to Combine

Now, apply the distributive property to combine the like terms. Combining \( 14 \sqrt{3} \) and \( 40 \sqrt{3} \), we factor out \( \sqrt{3} \): \( (14 + 40) \sqrt{3} = 54 \sqrt{3} \).

Key Concepts

Simplifying Radicals

In mathematics, simplifying radicals involves expressing a square root or any root in its simplest form. This is often done by breaking down the number under the root into its factors, aiming to isolate any perfect square factors. For example, when you see \( \sqrt{12} \), you can break it down as \( \sqrt{4 \times 3} \). Here, 4 is a perfect square, and its square root is 2. Therefore, we can simplify \( \sqrt{12} \) to \( 2 \sqrt{3} \). This technique is useful because it makes the expression easier to work with, especially in equations and simplifying expressions further.
Getting comfortable with identifying perfect squares is crucial for simplifying radicals. Some examples of perfect squares include 1, 4, 9, 16, 25, and so on. Whenever you see a number under a square root, think of its factorization to identify any of these perfect squares as factors.

Like Terms

Like terms in algebraic expressions are terms that contain the same variables raised to the same power, allowing them to be combined through addition or subtraction. When working with expressions that involve square roots, like terms have the same radicand (the number inside the root).
In our example, both \( 7 \sqrt{12} \) and \( 10 \sqrt{48} \) share the radicand of \( \sqrt{3} \) once simplified, becoming \( 14 \sqrt{3} \) and \( 40 \sqrt{3} \) respectively. Since they have the same radical part, they can be combined just like you would with simple algebra terms. Using the distributive property, you add or subtract only the coefficients of the like terms, treating the radical part as a common factor.
So to combine \( 14 \sqrt{3} \) and \( 40 \sqrt{3} \), calculate \( 14 + 40 \) to get 54. This results in \( 54 \sqrt{3} \), thus simplifying the expression while keeping the radical component the same.

Perfect Squares

Perfect squares play a pivotal role in simplifying radicals. They are integers which are squares of other integers. For instance, numbers like 4, 9, 16 are perfect squares because they are \( 2^2 \), \( 3^2 \), and \( 4^2 \) respectively. When simplifying radicals, we seek to identify these perfect square factors within a number under the radical sign to simplify the problem.
In the step-by-step solution of the example, \( \sqrt{12} \) becomes \( \sqrt{4 \times 3} \) and \( \sqrt{48} \) becomes \( \sqrt{16 \times 3} \). Here, 4 and 16 are perfect squares, which simplify to 2 and 4, respectively. By identifying and removing perfect square factors from a radical, you simplify the entire expression, making calculations easier and more efficient.
Understanding and quickly identifying perfect squares helps greatly in algebra as it makes simplification of expressions using radicals more straightforward. By isolating and simplifying using squares, complex scenarios become manageable.