Problem 2 - Algebra 2

Question

Simplify. $$ \sqrt{80} $$

Step-by-Step Solution

Answer

The simplified form of \(\sqrt{80}\) is \(4\sqrt{5}\).

Step 1: Understand the Problem

We need to simplify the expression \(\sqrt{80}\). Simplifying a square root involves reducing it by identifying and pulling out perfect square factors.

Step 2: Find Prime Factorization

Start by finding the prime factorization of 80. We know that \(80 = 8 \times 10\). Further breaking down, we get \(8 = 2^3\) and \(10 = 2 \times 5\). Hence, the prime factorization of 80 is \(2^4 \times 5\).

Step 3: Identify Perfect Squares

Now, with the prime factorization \(2^4 \times 5\), identify the perfect square(s). We notice \(2^4\) is a perfect square because \((2^2)^2 = 4^2\), and 4 is a perfect square. So, inside the square root, \(2^4\) can be written as \((2^2)^2\).

Step 4: Separate the Perfect Squares and Simplify

Write \(\sqrt{80}\) as \(\sqrt{(2^2)^2 \times 5}\). The square root of a product can be separated: \(\sqrt{(2^2)^2} \times \sqrt{5}\). Simplifying further, we get \(2^2 \times \sqrt{5}\), which is \(4\sqrt{5}\).

Step 5: Verify the Simplification

Verify by calculating \((4\sqrt{5})^2\) to check if it returns to 80. Calculate \((4\sqrt{5})^2 = 16 \times 5 = 80\), confirming that our simplification is correct.

Key Concepts

Prime Factorization

Prime factorization is essentially like breaking down a number into its building blocks. It means representing a number as a product of its prime numbers, which can help in various mathematical operations, including simplifying radicals. For example, when we talk about the prime factorization of 80, we start by recognizing that 80 can be divided into smaller components that are prime numbers.
Here is how you can do it:

• First, break 80 into 8 and 10, because 80 equals 8 times 10.
• Next, deconstruct 8 into 2 multiplied by 2 multiplied by 2, or expressed as \(2^3\).
• Then, 10 can be broken into 2 and 5, as 2 is a factor of 10 and 5 is a prime number.

When you combine these factors, the prime factorization of 80 is \(2^4 \times 5\). Essentially, you are dividing the number into its simplest prime components, making it easier to work with, especially when simplifying square roots.

Perfect Squares

Perfect squares are numbers that are the square of an integer. For example, 1, 4, 9, 16 are all perfect squares because they equal \(1^2\), \(2^2\), \(3^2\), and \(4^2\) respectively. Identifying perfect squares is crucial when simplifying radical expressions, as they allow us to reduce the terms inside a square root.
In the problem of simplifying \(\sqrt{80}\), after performing prime factorization and getting \(2^4 \times 5\), we find a perfect square among these factors. Here, \(2^4\) translates into \((2^2)^2\), which is effectively 16, a perfect square.
When simplifying radicals, pulling out the perfect square allows the expression within the square root to reduce significantly, leading to simplification. By recognizing and separating these perfect squares, you can transform the square root of a multiplication into a product of simpler square roots.

Square Roots

The square root of a number is one of two equal factors of a number. It answers the question: what number multiplied by itself will give the original number? When simplifying square roots, the goal is to express the root in its simplest form while maintaining its equality. This often involves using the previously discussed concepts of prime factorization and perfect squares.
In the context of \(\sqrt{80}\), once we identify that \(2^4\) is a perfect square, you can separate this from the radical expression. We express \(\sqrt{80}\) using its factors as \(\sqrt{(2^2)^2 \times 5}\), and further simplify it by breaking it into \(\sqrt{(2^2)^2} \times \sqrt{5}\).
Remember

• The square root of a perfect square results in an integer. In this case, \(\sqrt{(2^2)^2}\) simplifies to \(2^2\), which is 4.
• Leaves the \(\sqrt{5}\) as it is, as 5 is not a perfect square.

This results in the simplification of \(\sqrt{80}\) to \(4\sqrt{5}\), which is a much simpler expression and easier to manage in further calculations.