Problem 127 - Elementary Algebra

Question

Simplify. $$ \sqrt{\frac{28 q^{6}}{225}} $$

Step-by-Step Solution

Answer

\( \frac{2 q^3 \root{2}{7}}{15} \)

Step 1: Simplify the Radicand (Inside the Square Root)

The expression inside the square root is \( \frac{28 q^6}{225} \). First, factor the numerator and the denominator. Since 28 is \( 7 \times 4 \) and 225 is \( 15^2 \), we can rewrite the fraction inside the square root as \( \frac{7 \times 4 q^6}{15^2} \).

Step 2: Separate the Fraction

Use the property \( \frac{a}{b} = \frac{a}{1} \times \frac{1}{b} \) to separate the fraction: \( \frac{7 \times 4 \times q^6}{15^2} = 7 \times 4 \times \frac{q^6}{15^2} \).

Step 3: Distribute the Square Root

Apply the square root to the numerator and the denominator: \( \frac{\root{2}{28} \times \root{2}{q^6}}{\root{2}{225}} \).

Step 4: Simplify the Square Roots

Simplify each square root individually. \( \root{2}{28} = \root{2}{4 \times 7} = \root{2}{4} \times \root{2}{7} = 2 \times \root{2}{7} \) and \( \root{2}{q^6} = q^3 \), and \( \root{2}{15^2} = 15 \).

Step 5: Combine the Simplified Terms

Recombine the simplified terms: \( \frac{2 \times \root{2}{7} \times q^3}{15} = \frac{2 q^3 \root{2}{7}}{15} \). Thus, the simplified form is: \( \frac{2 q^3 \root{2}{7}}{15} \).

Key Concepts

Rationalizing the Denominator

When dealing with square roots and fractions, you'll often encounter situations where you need to rationalize the denominator. This means transforming the expression so that no radicals appear in the denominator.
For example, if you have \(\frac{1}{\root{2}{b}}\), you can rationalize the denominator by multiplying both the numerator and the denominator by \(\root{2}{b}\). This gives you \(\frac{\root{2}{b}}{b}\), which is much simpler to work with.
In our exercise, the denominator \(\root{2}{15^2}\) simplifies to 15. This step makes the final expression \(\frac{2 q^3 \root{2}{7}}{15}\) easier to interpret and handle in algebraic terms.

Factoring

Factoring is a key skill when simplifying square roots. It involves breaking down a number or expression into its constituent parts.
For instance, in the given exercise, the number 28 is factored into \(\root{2}{4 \times 7}\). Similarly, 225 is recognized as \(\root{2}{15^2}\). These factorizations help isolate the square root terms.
To simplify \(\root{2}{4 \times 7}\), we separate it into \(\root{2}{4} \times \root{2}{7}\), which simplifies further to \((2) \root{2}{7}\). Recognizing factorable components is crucial for simplifying complex expressions effectively.

Exponents

Understanding exponents is essential when working with algebraic expressions, especially involving roots. In our exercise, we have \(\root{2}{q^6}\).
Remember that a square root can be represented as an exponent: \(\root{2}{q^6} \) is the same as \((q^6)^{\frac{1}{2}}\). By applying the property of exponents, \((a^m)^n = a^{m \times n}\), we get \((q^6)^{\frac{1}{2}} = q^{6 \times \frac{1}{2}} = q^3\).
This simplification of exponents helps reduce complex expressions into more manageable terms, like the final result \(\frac{2 q^3 \root{2}{7}}{15}\). Mastering these exponent rules will boost your ability to handle various algebraic operations.