Problem 138 - Elementary Algebra

Question

Simplify. $$ \sqrt{\frac{75 r^{6} s^{8}}{48 r s^{4}}} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{5r^2 s^2 \sqrt{r}}{4}\).

Step 1: Simplify the fraction inside the square root

First, simplify the fraction inside the square root. The fraction is \ \ \ \ \(\frac{75 r^{6} s^{8}}{48 r s^{4}}\).\ \ \ \ Simplify the constants: \ \ \ \ \(\frac{75}{48}\) can be simplified. The greatest common divisor of 75 and 48 is 3. So, we get: \ \ \ \ \(\frac{75}{48} = \frac{25}{16}\)\ \ Next, simplify the variables. \ \ \ For \(r\): \(\frac{r^6}{r} = r^{5}\) \ \ \ For \(s\): \(\frac{s^8}{s^4} = s^{4}\)\ Thus, we have: \ \ \ \(\frac{75 r^{6} s^{8}}{48 r s^{4}} = \frac{25 r^{5} s^{4}}{16}\).

Step 2: Simplify inside the square root

Now that we have simplified the fraction, simplify the expression inside the square root:\ \ \ \(\sqrt{\frac{25 r^{5} s^{4}}{16}}\)

Step 3: Separate the square root

Separate the square root of the numerator and the square root of the denominator: \ \ \ \ \(\sqrt{\frac{25 r^{5} s^{4}}{16}} = \frac{\sqrt{25 r^{5} s^{4}}}{\sqrt{16}}\)

Step 4: Simplify the square root

Simplify the square roots: \ \ \ \ \(\sqrt{25} = 5\), \(\sqrt{r^{5}} = r^{2.5} = r^{2} \sqrt{r}\), \(\sqrt{s^{4}} = s^{2}\), and \(\sqrt{16} = 4\).\ \ \ Thus, we have: \ \ \ \ \(\frac{\sqrt{25 r^{5} s^{4}}}{\sqrt{16}} = \frac{5 r^{2} \sqrt{r} s^{2}}{4}\).

Key Concepts

fraction simplification

Fraction simplification involves making a fraction as simple as possible. This means reducing it to its lowest terms by dividing both the numerator and the denominator by their greatest common divisor (GCD).
In our exercise, we have the fraction \(\frac{75 r^{6} s^{8}}{48 r s^{4}}\). We simplify the constants first. The GCD of 75 and 48 is 3, so we divide both 75 and 48 by 3, getting \(\frac{25}{16}\).
Next, we simplify the variables. For the variable \(r\): \(\frac{r^6}{r} = r^{5}\). For \(s\): \(\frac{s^8}{s^4} = s^{4}\). Thus, after simplification, our fraction becomes \(\frac{25 r^{5} s^{4}}{16}\).
Remember, always look for the greatest common divisor and cancel out common factors in the numerator and denominator.

square roots

Square roots find the number that, when multiplied by itself, equals the given number. In algebra, to simplify expressions under a square root, we often use properties like \(\frac{\text{sqrt}(a)}{\text{sqrt}(b)} = \text{sqrt}(\frac{a}{b})\).
In our exercise, after simplifying the fraction, we have \(\frac{25 r^{5} s^{4}}{16}\). Under the square root, it becomes \(\text{sqrt}(\frac{25 r^{5} s^{4}}{16})\).
We can then separate the square root of the numerator and the denominator: \frac{\text{sqrt}(25 r^{5} s^{4})}{\text{sqrt}(16)}.
Breaking down the square roots: \(\text{sqrt}(25) = 5, \text{sqrt}(r^{5}) = r^{2.5} = r^{2} \text{sqrt}(r), \text{sqrt}(s^{4}) = s^{2}\), and \(\text{sqrt}(16) = 4\).
Hence, the simplified form is: \(\frac{5 r^{2} \text{sqrt}(r) s^{2}}{4}\).

variables simplification

Simplifying variables in fractions involves reducing powers of the same base by subtracting their exponents. This follows the rule \(\frac{a^m}{a^n} = a^{m-n}\).
In our problem, we look at \(\frac{r^{6}}{r}\) and \(\frac{s^{8}}{s^{4}}\). For \(r\), we subtract the exponents: \(\frac{r^{6}}{r} = r^{6-1} = r^{5}\). For \(s\), we get: \(\frac{s^{8}}{s^{4}} = s^{8-4} = s^{4}\).
Thus, our variables are simplified to \(\frac{25 r^{5} s^{4}}{16}\).
This rule makes simplification straightforward and helps in combining or breaking down algebraic expressions.

greatest common divisor

The greatest common divisor (GCD) is the largest number that divides both the numerator and the denominator of a fraction without leaving a remainder.
To simplify \(\frac{75}{48}\), we find their GCD, which is 3. Dividing 75 and 48 by 3, we get \(\frac{25}{16}\).
Steps to find the GCD might include:

• List out the factors of each number.
• Find the largest common factor.

This method is crucial for reducing fractions to their simplest form, making calculations easier and results more understandable.