Problem 41 - Algebra and Trigonometry

Question

Perform the indicated operations and simplify. $$ \left(\sqrt{a}-\frac{1}{b}\right)\left(\sqrt{a}+\frac{1}{b}\right) $$

Step-by-Step Solution

Answer

The simplified expression is \(a - \frac{1}{b^2}\).

Step 1: Identify the Expression Type

Notice that the expression given is in the form of \((x - y)(x + y)\), which is a difference of squares. In this case, \(x = \sqrt{a}\) and \(y = \frac{1}{b}\).

Step 2: Apply the Difference of Squares Formula

The formula for the difference of squares is \((x - y)(x + y) = x^2 - y^2\). Apply this to our expression, resulting in \((\sqrt{a})^2 - \left(\frac{1}{b}\right)^2\).

Step 3: Simplify Each Term

Calculate \((\sqrt{a})^2\), which simplifies to \(a\), and \((\frac{1}{b})^2\), which simplifies to \(\frac{1}{b^2}\).

Step 4: Write the Final Expression

Substitute the simplified terms back into the expression to get \(a - \frac{1}{b^2}\). This is the simplified form of the given expression.

Key Concepts

Difference of Squares

The concept of the difference of squares is a key algebraic identity that helps simplify expressions like the one in our exercise. This identity is expressed as \((x - y)(x + y) = x^2 - y^2\), and it holds true because the middle terms cancel each other out when the multiplication is carried out.
In our exercise, we have an expression that matches the format \((x - y)(x + y)\). By recognizing that \(x = \sqrt{a}\) and \(y = \frac{1}{b}\), we can directly apply this formula. The essential concept here is that instead of expanding using the distributive property, this shortcut allows for quicker simplification by simply squaring the two terms once the identity is recognized.
This identity is especially helpful in algebra for simplifying expressions because it reduces complex multiplication processes into simple subtraction.

Simplifying Expressions

Simplifying expressions involves reducing them to their most basic form without changing their value. In the context of the exercise, applying the difference of squares results in the expression \((\sqrt{a})^2 - \left(\frac{1}{b}\right)^2\). To simplify, it's essential to calculate each squared term individually:

• \((\sqrt{a})^2 = a\), because squaring a square root cancels out the square root sign.
• \(\left(\frac{1}{b}\right)^2 = \frac{1}{b^2}\), which simplifies the fraction to the numerator being squared and the denominator being multiplied by itself.

Substituting back these simplified forms gives us \(a - \frac{1}{b^2}\). This process of simplifying is crucial in algebra to make equations and expressions easier to work with by eliminating complex components.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols. They form the foundation of algebra and allow us to generalize arithmetic operations.
In the original exercise, the expression \(\left(\sqrt{a}-\frac{1}{b}\right)\left(\sqrt{a}+\frac{1}{b}\right)\) is an example of an algebraic expression. Here, the challenge is to manipulate and simplify the expression using algebraic techniques without changing its inherent value or meaning.
Understanding how to handle variables and operations is crucial for success in algebra. By recognizing patterns, like the difference of squares, students learn to transform complex expressions into simpler, more manageable forms. These skills are vital for solving more advanced algebraic equations and for applying mathematical reasoning in real-world contexts.