Problem 17 - Algebra 2 and Trigonometry

Question

In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ \sqrt{5 y} \cdot \sqrt{4 y^{3}} $$

Step-by-Step Solution

Answer

The simplest form is \(2y^2 \sqrt{5}\).

Step 1: Multiply the Radicands

When multiplying square roots, you can combine them under one square root by multiplying their radicands together. For this exercise, multiply the radicands: \(5y\) and \(4y^3\). So, \(\sqrt{5y} \cdot \sqrt{4y^3} = \sqrt{5y \cdot 4y^3} = \sqrt{20y^4}\).

Step 2: Simplify Inside the Square Root

Simplify the expression inside the square root. Start by multiplying the constants and then multiplying the variables:- The constants: \(5 \cdot 4 = 20\).- The variables: \(y \cdot y^3 = y^4\).Thus, you have \(\sqrt{20y^4}\).

Step 3: Simplify the Square Root

Simplify \(\sqrt{20y^4}\) by breaking it down:- \(20 = 4 \times 5\), and since \(\sqrt{4} = 2\), you can take \(2\) out of the square root.- \(y^4\) can be simplified as \((y^2)^2\), and \(\sqrt{(y^2)^2} = y^2\) because the square root and the square are inverse operations.Thus, \(\sqrt{20y^4} = \sqrt{4 \times 5} \cdot \sqrt{y^4} = 2y^2 \sqrt{5}\).

Key Concepts

Radicands Multiplication

To multiply radicands effectively, it's essential to understand that when you have two square roots multiplying each other, such as \( \sqrt{a} \cdot \sqrt{b} \), you can combine them inside one square root: \( \sqrt{ab} \). In our exercise, we apply this rule by combining \( \sqrt{5y} \) and \( \sqrt{4y^3} \).

• This becomes \( \sqrt{5y \cdot 4y^3} \).
• Multiply the numerical parts first: \(5 \cdot 4 = 20\).
• Then, multiply the variables: \(y \cdot y^3 = y^4\).

So our new expression under the square root is \( \sqrt{20y^4} \). This consolidation is the basis of simplifying radical expressions.

Square Root Simplification

Simplifying a square root means breaking it down to its simplest form. This can be done by separating the square root of perfect squares and leaving those that are not as they are. Take \( \sqrt{20y^4} \) from our exercise.

• The number 20 can be factored into \(4 \times 5\), where 4 is a perfect square.
• Since \( \sqrt{4} = 2\), we can simplify this part to 2.
• The variable part \(y^4\) can be seen as \((y^2)^2\).
• The square root of \((y^2)^2\) is \(y^2\) because the exponent and the square root cancel out.

Putting these simplifications together gives us \(2y^2\sqrt{5}\), where 5 stays under the square root because it's not a perfect square. This results in the simplest form for a non-negative radicand.

Variable Simplification

Often in simplifying expressions, especially with variables, it's crucial to deal wisely with powers and indices. From the exercise, \( \sqrt{y^4} \) is a critical component of simplification.

• The expression \(y^4\) within the square root can be rewritten as \((y^2)^2\).
• This manipulation allows us to apply the rule that the square root of \((x^2)\) is \(x\), which reduces \(\sqrt{(y^2)^2}\) to \(y^2\).

Handling variables inside square roots in this way simplifies potentially confusing terms, especially when they have high powers. It boils down to understanding that taking the square root of a square cancels out the power, making it a straightforward yet very powerful tool in algebra.