Problem 42 - Elementary Algebra

Question

For the following problems, simplify each expressions. $$ \frac{\sqrt{48 x^{6} y^{7}}}{\sqrt{3 x y}} $$

Step-by-Step Solution

Answer

Question: Simplify the expression \(\frac{\sqrt{48 x^{6} y^{7}}}{\sqrt{3 x y}}\). Answer: \(4x^{2}y^{2}\sqrt{y}\)

Step 1: Rewrite the expression in terms of square roots

Rewrite the expression as: $$ \frac{\sqrt{48 x^{6} y^{7}}}{\sqrt{3 x y}} = \frac{\sqrt{48}\,\sqrt{x^{6}}\,\sqrt{y^{7}}}{\sqrt{3}\,\sqrt{x}\,\sqrt{y}} $$

Step 2: Simplify the numeric part of the expression

Simplify the numeric part of the expression as follows: $$ \frac{\sqrt{48}}{\sqrt{3}} = \frac{\sqrt{16 \cdot 3}}{\sqrt{3}} = \frac{4\sqrt{3}}{\sqrt{3}}=\frac{4}{1}=4 $$

Step 3: Simplify the variable part of the expression

Simplify the variable part of the expression as follows: $$ \frac{\sqrt{x^{6}}}{\sqrt{x}} = \frac{x^{3}}{x^{1}} = x^{2} $$ and $$ \frac{\sqrt{y^{7}}}{\sqrt{y}} = \frac{y^{3}\sqrt{y}}{y^{1}} = y^{2}\sqrt{y} $$

Step 4: Combine the simplified terms

Combine the simplified terms to obtain the final expression: $$ 4x^{2}y^{2}\sqrt{y} $$ So, the simplified expression is \(4x^{2}y^{2}\sqrt{y}\).

Key Concepts

Radical Simplification

Understanding how to simplify radicals is crucial in algebra. When we talk about 'radical simplification', we are referring to the process of breaking down square roots (or higher roots) into the simplest possible form. The goal is to remove the radical symbol whenever possible, which can be done by finding square factors. For instance, consider the expression \(\sqrt{48}\). The number 48 can be divided into \(16 \times 3\), where 16 is a perfect square since \(4^2 = 16\). This means \(\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}\), simplifying the radical to \(4\sqrt{3}\). The same principle applies to variables with exponents under the radical. Variables with even exponents can be simplified since these exponents are squares of their roots. Therefore, \(x^6\) becomes \(x^{6/2} = x^3\), indicating that under the square root sometimes lies a perfect square that can be taken out of the radical, simplifying the expression.

Square Root Operations

Working with square root operations involves not only simplification but also knowing how to perform arithmetic with square root expressions. The initial step in your approach should be to separate numbers and variables under a common square root sign into individual square roots, like how we rewrote \(\sqrt{48 x^{6} y^{7}}\) as \(\sqrt{48}\sqrt{x^{6}}\sqrt{y^{7}}\).

Next, simplify each of these square roots separately if possible. Calculating the square root of an expression is essentially asking, 'What number multiplied by itself gives me this expression?' For numbers, you look for perfect squares as factors, while for variables, you adjust the exponent so that it is divided by 2. Afterward, combine the simplified components back together carefully. Keep in mind that when square roots are divided by each other, you can also simplify directly by dividing their insides, as we saw with \(\sqrt{48}/\sqrt{3}\) simplifying to \(\sqrt{16 \cdot 3}/\sqrt{3} = 4\).

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. They don't have an equal sign like an equation would. In our exercise, we see an example of an algebraic expression involving radicals and variables.

Simplifying the Algebraic Expression

The key to simplifying an algebraic expression with radicals, is to always first consider the 'like terms'. These are terms that have the same variables to corresponding powers. When simplifying \(\sqrt{x^{6}}/\sqrt{x}\), we consider the exponents of \(x\) and reduce them accordingly; similarly for \(\sqrt{y^{7}}/\sqrt{y}\). This step simplifies our algebraic expression down to \(4x^{2}y^{2}\sqrt{y}\), where we combined our like terms (the variables to the same power) in a simplified form.

Keeping track of all these features—how to break down and combine radicals and how to handle variables within them—is crucial for mastering algebra.