Problem 21 - College Algebra

Question

Use the quotient rule to simplify the expressions in Exercises \(17-26 .\) Assume that \(x>0\) $$\frac{\sqrt{48 x^{3}}}{\sqrt{3 x}}$$

Step-by-Step Solution

Answer

The simplified form of the given expression is \(4x\).

Step 1: Simplify the square root of \(48x^3\)

We first break down \(48x^3\) into prime factors to simplify it: \n \( 48x^3 = 2^4 * 3 * x^3 \).\nThen, we apply the square root: \n \( \sqrt{ 2^4 * 3 * x^3 } = 2^2*x \sqrt{3x} = 4x \sqrt{3x}\).

Step 2: Simplify the square root of \(3x\)

Now, remember the square root of \(3x\) is simply \(\sqrt{3x}\) . This can't be simplified any further.

Step 3: Apply the quotient rule

We now apply the quotient rule of radicals to simplify the entire fraction, i.e. divide the simplified square root of \( 48x^3 \) by the simplified square root of \(3x\): \n \( \frac{4x \sqrt{3x}}{\sqrt{3x}} \). Remember that anything divided by itself is 1, so we simplify to get 4x.

Key Concepts

Simplifying Square Roots

Breaking down square roots into simpler terms is an essential algebraic skill. When we talk about simplifying square roots, we're seeking to express a square root in its most reduced form. Take, for example, the square root of a number like 48. To simplify it, we look for perfect squares. A perfect square is any number that can be expressed as the square of an integer, like 4 or 16.
To simplify, we first factor 48 into its prime components. Prime factorization involves breaking down a number until all remaining factors are prime. For 48, this breakdown is:

• 48 = 2 × 2 × 2 × 2 × 3 (or 2^4 × 3).

Then, we apply the square root. The rule is that the square root of a product of factors is the product of the square roots of each factor. Thus we can write:

• \( \sqrt{48} = \sqrt{2^4 \times 3} = 2^2 \times \sqrt{3} = 4\sqrt{3} \).

This means the original \( \sqrt{48} \) can be simplified to \( 4\sqrt{3} \) by taking the square root of 16 (which is 4) and leaving \( \sqrt{3} \) as is. For algebraic terms like \( x^3 \), you apply similar logic by dividing the exponent by 2 if possible, simplifying it step by step.

Prime Factorization

Prime factorization is a method to express a number as a product of its prime numbers. It's a crucial concept to understand when simplifying radical expressions. To factorize, you start by dividing the number by the smallest prime, usually 2, and continue the process until all factors are prime.
Consider the original number 48. We break it down as:

• 48 divided by 2 gives 24,
• 24 divided by 2 gives 12,
• 12 divided by 2 gives 6,
• 6 divided by 2 gives 3, which is prime.

Thus, \( 48 = 2^4 \times 3 \). Similarly, for a variable with an exponent, like \( x^3 \), it is expressed as \( x \times x \times x \).
Using prime factorization, we can easily simplify expressions under radicals by pairing factors. For example, if \( x^3 \) is under a square root, you can simplify it to \( x\sqrt{x} \), because you take out one pair of x's. Understanding prime factorization helps you break down and simplify even complex algebraic expressions.

Radical Expressions

Radical expressions involve roots, most commonly square roots, and understanding their properties is key to simplifying them effectively. A radical expression can include numbers, variables, or a combination of both under a radical sign \( \sqrt{} \).
To simplify such expressions, you often use the quotient rule or the product rule. The quotient rule allows you to take the square root of a fraction by separating the numerator and the denominator under their own square roots. An example of the quotient rule in action is simplifying \( \frac{\sqrt{48x^3}}{\sqrt{3x}} \).
We simplify this by first addressing the entire fraction as \( \sqrt{\frac{48x^3}{3x}} \). Since the square root of a quotient is the quotient of the square roots, we express it as:

• \( \frac{\sqrt{48x^3}}{\sqrt{3x}} = \frac{4x\sqrt{3x}}{\sqrt{3x}} \).
• The \( \sqrt{3x} \) terms cancel out, leaving \( 4x \).

Recognizing how radical expressions behave with respect to the rules of arithmetic helps to simplify them correctly, leading to clarity in solving algebraic equations.