Problem 84 - Elementary Algebra

Question

Simplify. $$ \sqrt{175 n^{13}} $$

Step-by-Step Solution

Answer

5 n^6 \sqrt{7 n}

Step 1: Factorize the Radicand

Express 175 and the exponent of n in a simpler, factorized form. Note that 175 can be factorized as 25 * 7. So, rewrite the expression as \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[\sqrt{175 n^{13}} = \sqrt{25 \times 7 \times n^{13}}.\]

Step 2: Simplify Perfect Squares

Identify and separate the perfect square factors from the non-perfect square factors under the radical. We know that 25 is a perfect square, and that we can express \ n^{13} \ as \ n^{12} \ times \ n \. So we have: \ \ \ \[ \sqrt{25 \times 7 \times n^{13}} = \sqrt{25} \times \sqrt{7 n^{13}}. \] Since \ n^{12} \ is also a perfect square, it becomes \ (n^6)^2 \ within the radical, leading to: \ \[ \sqrt{25} \times \sqrt{7 n^{13}} = 5 \times \sqrt{(n^6)^2 n} = 5 n^6 \times \sqrt{7 n}. \]

Step 3: Combine the Simplified Terms

Multiply the simplified terms outside of the radical, and retain the radical term as needed. Therefore, the simplified form of the given expression is: \[ \sqrt{175 n^{13}} = 5 n^6 \sqrt{7 n}. \]

Key Concepts

Simplifying Radicals

When we simplify radicals, our goal is to make the expression under the radical as simple as possible. To achieve this, we look for perfect squares, perfect cubes, or any other perfect roots within the radicals.

For example, in this exercise, we have \(\sqrt{175 n^{13}}\). The first step is to factor the number and the variable inside the radical. We know that 175 can be factored into 25 and 7 since \(175 = 25 \times 7\). This helps us extract the perfect square, which in this case is 25:

• \(\sqrt{175 n^{13}} = \sqrt{25 \times 7 \times n^{13}}\)

Now, separate the perfect squares from the non-perfect squares. We know that \(25 = 5^2\), so using the rule \(\textrm{\sqrt{a^2} = a}\), we get:

• \(\sqrt{25} = 5\)

Therefore, we can simplify \(\sqrt{25} \times \sqrt{7 n^{13}}\) to:

• \(\sqrt{25} \times \sqrt{7 n^{13}} = 5 \sqrt{7 n^{13}}\)

Factoring

Factoring plays a crucial role in simplifying expressions, especially when dealing with radicals. By breaking numbers and expressions into their factors, we can identify and simplify parts of them more easily.

In our example, we need to split \(175\) and \(n^{13}\) into their factors. We factorize 175 into 25 and 7. However, for the variable part \(n^{13}\), we express it as a product involving a perfect square:

• \(\sqrt{25 \times 7 \times n^{13}}\)
• \(n^{13} = n^{12} \times n\)

We write \(n^{13}\) as \(n^{12} \times n,\) where \(n^{12}\) is a perfect square (\(n^6)^2\):

Thus the expression becomes:

• \(\sqrt{25 \times 7 \times n^{13}} = \sqrt{25} \times \sqrt{7 n^{13}}\)

In this form, we can separate the parts involving perfect squares, leading to a simpler expression.

Exponents and Powers

Understanding exponents and powers is critical for manipulating expressions and simplifying them. When we encounter terms like \(n^{13}\), we need to break them into smaller parts that are easier to work with.

In the given problem, \(n^{13}\) isn't easy to simplify directly under the radical because its exponent isn't a perfect square. However, by expressing \(n^{13}\) as \(n^{12} \times n,\) we split it into a part that's a perfect square \( (n^6)^2\) and another simple term \(n.\)

• \(\textrm{n^{13}} = n^{12} \times n\)

Now apply the product rule of radicals which states:

• \(\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}\)

This allows us to separate and simplify the expression further:

• \(\sqrt{25} \times \sqrt{7 (n^6)^2 n} = 5 \cdot n^6 \cdot \sqrt{7 n} \)

Thus, using our knowledge of exponents and powers, we transform \(n^{13}\) to a simpler form, resulting in:

• \(\sqrt{175 n^{13}} = 5 n^6 \sqrt{7 n}\)