Problem 71 - College Algebra with Corequisite Support

Question

For the following exercises, simplify each expression. \(\frac{x \sqrt{64 y}+4 \sqrt{y}}{\sqrt{128 y}}\)

Step-by-Step Solution

Answer

\( \frac{(2x + 1)\sqrt{2}}{4} \)

Step 1: Simplify Inside the Square Roots

First, let's simplify the terms inside the square roots. We know that \(\sqrt{64} = 8\) because \(64\) is a perfect square. Similarly, \(\sqrt{128}\) can be simplified further. We break it down: \(\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}\).

Step 2: Substitute Simplified Roots

Based on the first step, we replace \(\sqrt{64}\) and \(\sqrt{128}\) in the original expression. This gives us: \[\frac{x \cdot 8 \cdot \sqrt{y} + 4 \sqrt{y}}{8 \sqrt{2} \cdot \sqrt{y}}.\]

Step 3: Simplify the Expression

Factor out a common factor from the numerator. The numerator is now \(\sqrt{y} (8x + 4)\). So the expression becomes: \[\frac{\sqrt{y} (8x + 4)}{8 \sqrt{2} \cdot \sqrt{y}}.\]

Step 4: Cancel the Common Factors

Now, you can cancel \(\sqrt{y}\) from the numerator and the denominator since they are common factors. This simplifies the expression to: \[\frac{8x + 4}{8\sqrt{2}}.\]

Step 5: Simplify the Remaining Fraction

Notice the common factor \(4\) in \(8x + 4\), hence we factor out \(4\):\[\frac{4 (2x + 1)}{8\sqrt{2}}.\]Now simplify the fraction by dividing both the numerator and the denominator by \(4\):\[\frac{2x + 1}{2\sqrt{2}}.\]

Step 6: Rationalize the Denominator

To eliminate the square root from the denominator, multiply the numerator and the denominator by \(\sqrt{2}\). This gives us: \[\frac{(2x + 1)\sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{(2x + 1)\sqrt{2}}{4}.\]

Key Concepts

Square Roots Simplification

Simplifying square roots can make complex expressions more manageable. A square root is generally simplified by removing perfect square factors.
For instance, let's look at \(\sqrt{64}\), which is easy because 64 is a perfect square, i.e., \(\sqrt{64} = 8\). If the square root is not initially obvious, we should factor it. Take \(\sqrt{128}\) as an example: \(\sqrt{128} = \sqrt{64 \times 2}\).
Since \(64\) is a perfect square, it becomes \(8\) and you end with \(8\sqrt{2}\).

• Identify perfect squares within the number.
• Extract these squares as a factor outside the root.
• Keep non-perfect squares inside the root.

By simplifying square roots, you untangle expressions, making them much simpler to work with.

Rationalizing the Denominator

Having a square root in the denominator can be cumbersome. To clear it, we rationalize the denominator.
This means converting the denominator to a rational number (without a root), which is done by multiplying both the top and the bottom of the fraction by a suitable term.
In our exercise, we ended with the expression \(\frac{(2x + 1)}{2\sqrt{2}}\).
You can remove the root in the denominator by multiplying the whole fraction by \(\sqrt{2}/\sqrt{2}\), resulting in \(\frac{(2x + 1)\sqrt{2}}{2\sqrt{2} \times \sqrt{2}}\=\frac{(2x + 1)\sqrt{2}}{4}\).

• Multiply both numerator and denominator by the square root present in the denominator.
• Ensure the result keeps the expression equivalent in value.

Rationalizing makes expressions cleaner and allows for easier arithmetic operations.

Factoring in Algebra

Factoring simplifies expressions by breaking them into products of smaller expressions. It's a valuable tool in algebra for simplifying, solving equations, and more.
In our problem, we noticed a common factor in the expression \(8x + 4\). The greatest common factor (GCF) here is \(4\), so we factor it out:
\((8x + 4) = 4(2x + 1)\). This leaves us with a simpler term to work with.

• Identify the GCF in your terms.
• Divide each term by this common factor.
• Rewrite the expression as a product of the GCF and the simplified terms.

Factoring helps reduce complexity in expressions and is essential for solving many algebraic equations effectively.