Problem 72 - Elementary Algebra

Question

Simplify each expression by performing the indicated operation. $$ \frac{1}{3-\sqrt{2}} $$

Step-by-Step Solution

Answer

Answer: \(\frac{3+\sqrt{2}}{7}\)

Step 1: Identify the conjugate of the denominator

We need to find the conjugate of the denominator, which has the form \({(a+\sqrt{b})}\) if the original denominator is \({(a-\sqrt{b})}\). In our case, the denominator is \({(3-\sqrt{2})}\), so the conjugate is \({(3+\sqrt{2})}\).

Step 2: Multiply the numerator and denominator by the conjugate

To eliminate the square root from the denominator, we'll multiply both the numerator and denominator by the conjugate we found in Step 1: \({(3+\sqrt{2})}\). This gives us: $$\frac{1}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}}$$

Step 3: Apply the difference of squares formula

Now we can apply the difference of squares formula, which is \({(a-b)(a+b)=a^2-b^2}\). In our case, we'll apply it to the denominator: $$(3-\sqrt{2})(3+\sqrt{2}) = 3^2 - (\sqrt{2})^2$$

Step 4: Simplify the expression

Now let's simplify the numerator and the denominator: Numerator: \(1 \times (3+\sqrt{2})=(3+\sqrt{2})\) Denominator: \(3^2 - (\sqrt{2})^2 = 9-2 = 7\) Putting it all together, our simplified expression is: $$\frac{3+\sqrt{2}}{7}$$

Key Concepts

Conjugate of a Binomial

Understanding the 'conjugate of a binomial' is a stepping stone in the process of simplifying radical expressions. A binomial conjugate involves changing the sign between two terms in a binomial. For instance, if you have a binomial such as \((a - b)\), its conjugate would be \((a + b)\).

Why is this useful? When you multiply a binomial by its conjugate, the result is the 'difference of squares', which is a fundamental algebraic identity that helps to simplify expressions, especially when rationalizing the denominator (which we'll discuss later).

Why Use Conjugates?

Rationalizing the denominator often requires the use of the conjugate to eliminate radicals from the bottom of a fraction. By multiplying the numerator and denominator by the binomial's conjugate, you maintain the equivalence of the expression while creating a scenario where the denominator becomes a whole number.

Difference of Squares

The 'difference of squares' is an algebraic pattern that you'll see frequently when simplifying expressions with radicals. It states that for any two terms \(a\) and \(b\), the product of their sum and difference will be the difference of their squares: \((a+b)(a-b) = a^2 - b^2\).

This principle is incredibly helpful when attempting to rationalize a denominator containing a square root. By multiplying by the appropriate conjugate, you create a difference of squares in the denominator, which results in a perfect square - and, subsequently, an integer - thus making the expression more straightforward.

Application in Exercises

In the context of the exercise given, after multiplying by the conjugate, the denominator becomes \(3^2 - (\sqrt{2})^2\), which simplifies to 7, a rational number. This use of the difference of squares property is essential in simplifying radical expressions.

Rationalizing the Denominator

The process of 'rationalizing the denominator' is performed to remove radicals from the denominator of a fraction. In basic terms, it's about making the denominator rational. To do this, you can multiply both the numerator and the denominator of the fraction by the conjugate of the denominator's binomial.

Rationalization is especially important in mathematical contexts where you need precise calculations without the complications introduced by radicals. It’s also a convention in mathematics to present expressions in their simplest form.

Why Rationalize?

Having a non-rational denominator can make further algebraic manipulation cumbersome. Rationalizing the denominator simplifies the expression and makes it easier to work with, especially when adding, subtracting, or comparing fractions.

Elementary Algebra

At the core of manipulating and simplifying expressions in mathematics is 'elementary algebra'. This field of mathematics is about understanding and applying the rules and operations that allow us to solve equations and simplify expressions involving variables and constants.

From solving for unknowns to working with polynomials, elementary algebra plays a crucial role in building a strong foundation for higher-level math. It is essential in teaching us how to handle abstract concepts and apply them to various problems—like rationalizing denominators or finding conjugates.

Practical Uses

Elementary algebra is not just an academic exercise. It equips students with problem-solving tools that are used in various fields, from engineering and science to finance and data analysis. It's all about finding patterns, simplifying complex ideas, and making logical deductions, which are valuable skills in any discipline.