Problem 10 - Algebra 2 and Trigonometry

Question

In \(3-38\) write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero. $$ 8 \sqrt{11 b^{4}}-\sqrt{99 b^{4}} $$

Step-by-Step Solution

Answer

The simplest form of the expression is \(5b^2 \sqrt{11}\).

Step 1: Simplify Each Radical

Let's simplify each square root separately. The expression is \(8 \sqrt{11b^4} - \sqrt{99b^4}\). For \(8 \sqrt{11b^4}\), we note that \(\sqrt{b^4} = b^2\) since \(b\) is non-negative. So, it becomes \(8b^2 \sqrt{11}\). Similarly, for \(\sqrt{99b^4}\), it becomes \(b^2 \sqrt{99}\). Thus, the expression is \(8b^2 \sqrt{11} - b^2 \sqrt{99}\).

Step 2: Simplify the Second Radical Further

We need to break down \(\sqrt{99}\). Notice that \(99 = 9 \times 11\), and therefore, \(\sqrt{99} = \sqrt{9 \times 11} = \sqrt{9} \times \sqrt{11} = 3 \sqrt{11}\). The expression becomes \(8b^2 \sqrt{11} - b^2 (3 \sqrt{11})\).

Step 3: Factor Out Common Terms

Now, factor \(b^2 \sqrt{11}\) from the expression: \(8b^2 \sqrt{11} - 3b^2 \sqrt{11} = (8 - 3)b^2 \sqrt{11}\). This simplifies to \(5b^2 \sqrt{11}\).

Key Concepts

Square Roots

Square roots are fundamental in simplifying radicals. They allow us to find a number that, when multiplied by itself, results in the original number inside the radical symbol. In algebraic expressions involving square roots, such as \( \sqrt{11b^4} \), it's essential to simplify wherever possible to deal with the expression efficiently.
When addressing variables under the square root, like \( b^4 \), it's vital to note that \( \sqrt{b^4} = b^2 \). Here’s why:

• Squaring a non-negative variable like \( b \) twice (since \( b^2 \times b^2 = b^4 \)) allows us to pull \( b^2 \) out of the square root easily.
• This simplification means we are left with the simpler expression \( b^2 \sqrt{11} \), which is easier to work with.

Understanding these basic square root rules is a stepping stone to handling more complex algebraic expressions with radicals.

Factoring Radicals

Factoring radicals is about breaking down the number inside the square root to simplify it as much as possible. For instance, with \( \sqrt{99} \), we find factors to make simplification easier.
Here's how we did it:

• Identify that \( 99 \) can be factored into \( 9 \times 11 \).
• Since \( 9 \) is a perfect square, its square root is \( 3 \) (because \( 3 \times 3 = 9 \)).

Thus, \( \sqrt{99} = 3 \sqrt{11} \). This reduces computations in further algebraic manipulations.
By breaking down the radical into its prime factors or known perfect squares, you simplify the form, just as in the original exercise where \( 99 \) was decomposed to make multiplication and subtraction easier.

Algebraic Expressions

Manipulating algebraic expressions involving radicals requires careful attention to detail to maintain their equivalent forms. In our exercise, this involved expressions like \( 8b^2 \sqrt{11} - 3b^2 \sqrt{11} \).
To simplify these expressions:

• Look for common factors. In this case, \( b^2 \sqrt{11} \) is a shared factor in both terms, so factoring it out is logical.
• Once factored, you simplify: \( (8 - 3) b^2 \sqrt{11} \), which gives \( 5b^2 \sqrt{11} \).

By consistently applying algebraic principles such as factoring, you'll make expressions more manageable and much easier to interpret.
This skill is central in algebra, aiding in solving equations and understanding the relationships between variables.