Problem 70 - Algebra 2

Question

Simplify each expression. $$ \frac{\sqrt{64 x^{4}}}{\sqrt{144 x^{5}}} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{2}{3x^{0.5}}\).

Step 1: Simplifying the Radicals

We'll start by simplifying each radical fraction separately. The square root of \(64x^4\) is \(8x^2\), and the square root of \(144x^5\) is \(12x^{2.5}\). Hence, the expression becomes: \[\frac{8x^2}{12x^{2.5}}\]

Step 2: Rationalize the denominator and simplify

Rationalize the denominator by multiplying both the numerator and the denominator by \(x^{0.5}\). The expression, therefore, becomes: \[\frac{8x^2.5}{12x^{3}}\] Then simplify by dividing the top and bottom by \(4x^2.5\): \[\frac{2}{3x^{0.5}}\]

Key Concepts

square roots

A square root is a value that, when multiplied by itself, gives the original number. For example, the square root of 64 is 8 because 8 times 8 equals 64. The square root symbol is written as \( \sqrt{} \).
When simplifying square roots with variables, like \( \sqrt{64x^4} \), we treat numbers and variables separately. Here, \( \sqrt{64} = 8 \) because the square root of 64 is 8, and \( \sqrt{x^4} = x^2 \) because multiplying \( x^2 \) by itself gives \( x^4 \).
This means \( \sqrt{64x^4} = 8x^2 \).

• For even powers of a variable, divide the power by two to find the square root (e.g., \( x^4 \to x^2 \)).
• If the power is odd, split it into a part that can be squared (if possible) and a remainder (e.g., \( x^5 = x^4 \times x \)).

rationalizing the denominator

Rationalizing the denominator involves eliminating roots from the bottom of a fraction. This process makes the expression simpler and easier to understand. We commonly do this when we have a square root in the denominator.
For example, if we have \( \frac{1}{\sqrt{2}} \), we multiply both the numerator and the denominator by \( \sqrt{2} \) to eliminate the root from the bottom:\[ \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]
In our exercise, we have \( \frac{8x^2}{12x^{2.5}} \), and to rationalize \( x^{2.5} \) in the denominator, we multiply by \( x^{0.5} \). Doing so will give each term a whole exponent:\[ \frac{8x^{2.5}}{12x^{3}} \]This removes the fractional exponent from the denominator, making the expression clearer. Rationalizing is always about getting rid of square roots or fractional exponents in the denominator.

algebraic simplification

Algebraic simplification is the process of making an expression easier to work by combining and reducing terms. This includes operations like canceling common factors in fractions, combining like terms, and simplifying complex expressions.
In our problem, after rationalizing, we simplify \( \frac{8x^{2.5}}{12x^3} \) by dividing both the numerator and the denominator by their greatest common factor, which is \( 4x^{2.5} \).
This yields the simplified expression:\[ \frac{2}{3x^{0.5}} \]
Some key steps to simplifying algebraic expressions include:

• Identify and divide by common factors.
• Combine like terms, which are terms with the same variables and powers.
• Be mindful of negative exponents, as they represent reciprocal powers (e.g., \( x^{-1} = \frac{1}{x} \)).

exponents and powers

Exponents and powers tell us how many times to multiply a number by itself. For example, \( x^4 \) means \( x \times x \times x \times x \).
Understanding exponents is crucial when simplifying expressions, as they help indicate the relationship and operations to perform.
In our problem, we encounter expressions like \( x^{2.5} \) or \( x^{3} \). Exponents can be whole numbers, fractions, or negative numbers, and they follow specific rules:

• Product of Powers: \( x^a \times x^b = x^{a+b} \)
• Quotient of Powers: \( \frac{x^a}{x^b} = x^{a-b} \)
• Power of a Power: \( (x^a)^b = x^{ab} \)
• Negative Exponent: \( x^{-a} = \frac{1}{x^a} \)

The fractional exponent \( x^{0.5} \) is an extension of these rules and represents the square root \( \sqrt{x} \). So in our simplified expression \( \frac{2}{3x^{0.5}} \), \( x^{0.5} \) is equivalent to having a square root in the denominator.