Problem 193 - Elementary Algebra

Question

In the following exercises, simplify. $$ \frac{1}{6} \sqrt{27}-\frac{3}{8} \sqrt{48} $$

Step-by-Step Solution

Answer

-\sqrt{3}

Step 1: Simplify the Square Roots

First, break down the numbers inside the square roots into their prime factors to simplify them. \ \( 27 = 3 \times 3 \times 3 = 3^{3} \), so \( \ \sqrt{27} = \ \sqrt{3^{3}} = 3 \sqrt{3} \) \ \( 48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^{4} \times 3 \), so \ \( \sqrt{48} = \sqrt{2^{4} \times 3} = 2^{2} \times \ \sqrt{3} = 4 \sqrt{3} \)

Step 2: Substitute Simplified Roots Back

Substitute the simplified versions of the square roots back into the expression: \ \( \ \frac{1}{6} \sqrt{27} - \frac{3}{8} \sqrt{48} = \frac{1}{6} (3\sqrt{3}) - \frac{3}{8} (4\sqrt{3}) \ \)

Step 3: Simplify the Expression

Multiply the constants by the square root of 3: \( \frac{1}{6} \times 3 \sqrt{3} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \) \( \frac{3}{8} \times 4 \sqrt{3} = \frac{12\sqrt{3}}{8} = \frac{3\sqrt{3}}{2} \)

Step 4: Perform the Subtraction

Now, subtract the two terms: \ \( \frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = \frac{\sqrt{3}- 3\sqrt{3}}{2} = \frac{-2\sqrt{3}}{2} = -\sqrt{3} \)

Key Concepts

Square Roots

Square roots can seem tricky, but they play a fundamental role in simplifying algebraic expressions. When you take the square root of a number, you're trying to find another number which, when multiplied by itself, gives you the original number. For instance, \( \sqrt{9} = 3 \) because \(3 \times 3 = 9\).
Squares and roots go hand in hand in algebra. Understanding prime factorization (which we will discuss next) can greatly improve your ability to simplify square roots. Breaking down numbers into their prime factors makes it easier to simplify square roots by finding pairs of factors.

Prime Factorization

Prime factorization involves breaking down a number into its prime factors. A prime number is a number greater than 1 that has no divisors other than 1 and itself. For example, the number 27 can be broken down into \([3 \times 3 \times 3] \). This is written as \3^3\, indicating it is composed of prime number 3 repeated three times.
Similarly, 48 can be factored into \(2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3\). By breaking numbers into their prime factors, it is easier to identify pairs under the square root symbol.
After prime factorization, it becomes simple to simplify the square roots by grouping these prime factors into pairs.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form. Once the square roots are simplified using their prime factors, those simplified values are substituted back into the expression.
For instance, \( \sqrt{27} = 3 \sqrt{3} \) and \( \sqrt{48} = 4 \sqrt{3} \). Substituting in an expression like \( \frac{1}{6} \sqrt{27} - \frac{3}{8} \sqrt{48}\), becomes \( \frac{1}{6} (3 \sqrt{3}) - \frac{3}{8} (4 \sqrt{3})\).
Simplifying further, one gets \( \frac{3\sqrt{3}}{6} - \frac{12\sqrt {3}}{8} = \frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \).
Both terms are now simplified further by performing the subtraction, leading to the final solution.

Subtraction of Radicals

Radicals, or square roots in most cases, follow similar rules to variables when performing operations like addition or subtraction. To subtract radicals, ensure they have the same radicand (the number inside the square root).
In this problem, after simplifying, both terms under subtraction are \( \sqrt{3}\frac}{\underline{\phantom{xx}}} \). This allows us to subtract directly: \( \frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}= \frac{\sqrt{3} - 3\sqrt{3}}{2} = \frac{-2\sqrt{3}}{2} = -\sqrt{3}\).
Notice, the coefficients of \(\sqrt{3}\) are subtracted just like with simple algebraic terms, resulting in a solution where radicals are managed with straightforward arithmetic once they're simplified.