Problem 79 - Algebra for College Students

Question

How would you simplify the expression \(\frac{\sqrt{8}+\sqrt{12}}{\sqrt{2}}\) ?

Step-by-Step Solution

Answer

The simplified expression is \(2 + 2\sqrt{\frac{3}{2}}\).

Step 1: Simplify the square roots in the numerator

First, simplify the individual square roots in the numerator, \(\sqrt{8}\) and \(\sqrt{12}\).For \(\sqrt{8}\), factorize it into \(\sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\).For \(\sqrt{12}\), factorize it into \(\sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\).Thus, the numerator \(\sqrt{8} + \sqrt{12}\) becomes \(2\sqrt{2} + 2\sqrt{3}\).

Step 2: Factor out the common terms

Identify and factor out the common term in the expression \(2\sqrt{2} + 2\sqrt{3}\).Both terms have a common factor of \(2\), so factor it out: \(2(\sqrt{2} + \sqrt{3})\).

Step 3: Simplify the fraction

Substitute the factored form back into the expression: \(\frac{2(\sqrt{2} + \sqrt{3})}{\sqrt{2}}\).Now, divide each term in the parentheses by \(\sqrt{2}\):\[\frac{2\sqrt{2}}{\sqrt{2}} + \frac{2\sqrt{3}}{\sqrt{2}} = 2 + 2\frac{\sqrt{3}}{\sqrt{2}}\].

Step 4: Simplify the individual terms

Simplify each individual term.The first term, \(2\), remains as it is because \(\frac{2\sqrt{2}}{\sqrt{2}} = 2\).The second term, \(2\frac{\sqrt{3}}{\sqrt{2}}\), can be rewritten using the property \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\). Thus, it becomes \(2\sqrt{\frac{3}{2}}\).

Step 5: Write the final simplified expression

Combine the simplified terms:The expression \(2 + 2\sqrt{\frac{3}{2}}\) cannot be simplified further, thus it is the final simplified form of the original statement.

Key Concepts

Square Roots

Square roots are used to find the number that, when multiplied by itself, gives the original number. The square root function is often seen with a radical symbol (√).
To simplify a square root, you look for factors of the number that are perfect squares.
For instance, with \(\sqrt{8}\), we can break it down into \(\sqrt{4 \times 2}\). Since \(\sqrt{4}\) equals 2, this simplifies to \(2\sqrt{2}\). The process helps to streamline expressions and make them easier to work with.
Whenever possible, simplifying square roots can reduce complexity, making further calculations more straightforward.

Factoring

Factoring involves breaking down numbers or expressions into products of simpler numbers or factors.
In algebra, we often want to find common factors to help simplify expressions. Let's take \(2\sqrt{2} + 2\sqrt{3}\) as an example. Both terms share a common factor of 2.
By factoring 2 out, the expression becomes \(2(\sqrt{2} + \sqrt{3})\).
Factoring can greatly simplify algebraic expressions. It involves recognizing patterns or commonalities within terms, making complex expressions much more manageable.

Rationalizing Denominators

Rationalizing a denominator involves eliminating radicals from the denominator of a fraction. This process often makes expressions easier to handle, especially in further operations or functions.
When we have \(\frac{2\sqrt{3}}{\sqrt{2}}\), we can simplify it using the property of combining square roots: \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\).
By rewriting the expression as \(2\sqrt{\frac{3}{2}}\), the radical is now part of the numerator, resulting in a rationalized denominator.
This technique ensures that expressions maintain a cleaner and often more useful form, particularly in expressions requiring further simplification or evaluation.

Algebraic Fractions

Algebraic fractions involve expressions in the form of fractions where the numerator and/or the denominator are algebraic expressions.
When working with algebraic fractions, it's important to simplify wherever possible.
The fraction \(\frac{2(\sqrt{2} + \sqrt{3})}{\sqrt{2}}\) can be simplified by dividing each term in the numerator by \(\sqrt{2}\).
This yields \(2 + 2\frac{\sqrt{3}}{\sqrt{2}}\), an expression easier to handle and understand.
Thus, simplifying algebraic fractions can unveil more inherent properties and interrelationships of the involved expressions.