Problem 45 - College Algebra

Question

Simplify each expression. $$\sqrt{50 y^{8}}$$

Step-by-Step Solution

Answer

The simplified expression is \( 5y^4\sqrt{2} \).

Step 1: Factor Inside the Square Root

We start by factoring the expression inside the square root. Notice that 50 can be factored into prime factors as \(50 = 2 imes 25 = 2 imes 5^2\). The expression becomes: \[\sqrt{50y^8} = \sqrt{2 imes 5^2 imes y^8}\]

Step 2: Apply the Square Root Property

The square root property tells us that \(\sqrt{a imes b} = \sqrt{a} \times \sqrt{b}\). Applying this property, we separate the terms inside the square root:\[\sqrt{2 imes 5^2 imes y^8} = \sqrt{2} imes \sqrt{5^2} imes \sqrt{y^8}\]

Step 3: Simplify Each Part

Now, simplify each of the square roots individually:\(\sqrt{5^2} = 5\) because the square root of a square is the base itself.\(\sqrt{y^8} = y^4\) because taking the square root of a power is equivalent to dividing the exponent by 2.Therefore, the expression becomes:\[5y^4\sqrt{2}\]

Step 4: Combine the Results

Combine the simplified parts to express the simplified form of the original expression:\[5y^4\sqrt{2}\] This is the fully simplified form of \( \sqrt{50y^8} \).

Key Concepts

Square Root Property

The square root property is a valuable tool when simplifying expressions under a radical. It states that for any two positive numbers, \(a\) and \(b\), the square root of their product can be separated into the product of their individual square roots. This idea is captured as:

⦁ \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\)

This property allows us to break down complex expressions into simpler units, making them easier to simplify.

For instance, in our problem where we encounter \(\sqrt{50y^8}\), this property helps us to deal with parts separately. We can manage numbers and variables distinctly, applying the square root to each and re-combining them in the end.

Exponent Rules

Understanding exponent rules is essential in simplifying algebraic expressions, particularly those involving radicals. One fundamental rule when dealing with exponents is the power of a power rule, which helps when taking square roots of variables raised to a power.

⦁ The rule is: \((x^m)^n = x^{mn}\).

An important consequence of this rule is that when you take the square root of a variable raised to an exponent, it's akin to dividing the exponent by 2.

For example, in the expression \(\sqrt{y^8}\), applying this rule means we find \(y^{8/2} = y^4\). This demonstrates how exponent rules make simplifying radicals more manageable. Mastery of these rules makes working with complex algebraic expressions more intuitive.

Prime Factorization

Prime factorization involves breaking down a number into its prime factors. This technique is invaluable when simplifying expressions inside a square root, as it lays the foundation for further simplification.

For example, the number 50 can be factorized as \(50 = 2 \times 5^2\). By expressing numbers as products of primes, we can efficiently apply the square root property.

Identifying prime factors allows us to recognize perfect squares within an expression. In our case, the factor \(5^2\) is a perfect square, simplifying significantly when isolated in a square root.

Prime factorization turns otherwise challenging problems into a series of simple arithmetic steps, crucial for successful algebraic manipulation.

Radicals

Radicals often represent roots, mostly square roots, in mathematical expressions. Simplifying expressions with radicals is a skill that combines various techniques, such as utilizing properties of square roots, exponent rules, and prime factorization.

For example, in an expression like \(\sqrt{50y^8}\), different parts require different treatments:

• The number 50 needs prime factorization to identify any squares.
• The variable \(y^8\) requires exponent rules to simplify the expression efficiently.

By breaking down the radical into components, we apply specific simplification techniques, eventually recombining them into the simplest form. The result of the entire simplification process is \(5y^4\sqrt{2}\), showcasing the power of applying all these concepts effectively.