Problem 47 - Algebra 1: Concepts and Skills

Question

Simplify the radical expression. $$ \frac{6}{6+\sqrt{3}} $$

Step-by-Step Solution

Answer

The simplified form of the given expression is \( 1.0909(1.0909 - 0.1818\sqrt{3})\)

Step 1: Identify the conjugate of the denominator

The conjugate of \(6+\sqrt{3}\) is \(6-\sqrt{3}\). The conjugate of a binomial is obtained by changing the sign of the second term.

Step 2: Multiply by the conjugate

Multiply both the numerator and the denominator by the conjugate of the denominator: \((6)/(6+\sqrt{3}) * (6-\sqrt{3})/(6-\sqrt{3})\) This results to:\( (6 * (6 - \sqrt{3})) / ((6+\sqrt{3})(6-\sqrt{3})) \)

Step 3: Expand the expressions

Expand both the numerator and the denominator:\( (36 - 6\sqrt{3}) / ((6*6)-(sqrt{3}*sqrt{3}) \)Which simplifies to:\( (36 - 6\sqrt{3}) / ((36 - 3))\)

Step 4: Simplify the expression

By simplifying the expression, we get:\( (36 - 6\sqrt{3}) / (33) \)Or to put it another way:\( 1.0909(1.0909 - 0.1818\sqrt{3})\)

Key Concepts

Conjugate in Simplifying Radicals

When simplifying radical expressions, especially those with square roots in the denominator, the conjugate is extremely helpful. In mathematics, the conjugate of a binomial expression involves changing the sign between the two terms. For example, if you have a binomial like \(a + b\), its conjugate would be \(a - b\). This technique is particularly useful because when you multiply a binomial by its conjugate, you eliminate the middle term, resulting in a simple expression without any radicals.

The main reason for using the conjugate in radical expressions is to clear the denominator of any radicals. Having a radical in the denominator could make further mathematical operations challenging and, at times, incorrect. By multiplying by the conjugate, you replace the radical with an easier-to-handle term, usually a rational number. In our exercise, dealing with \(6 + \sqrt{3}\), its conjugate is \(6 - \sqrt{3}\). This allows us to remove the root in the denominator when we simplify the expression.

Rationalizing the Denominator

Rationalizing the denominator is a process used to eliminate radicals from the bottom of a fraction. For a clear and standardized answer, it is common practice to avoid leaving radicals in the denominator. This is why rationalizing plays a significant role in simplifying expressions that contain square roots or other irrational terms.

To rationalize the denominator of an expression like \(\frac{6}{6+\sqrt{3}}\), you multiply both the numerator and the denominator by the conjugate of the denominator, which is \(6-\sqrt{3}\). Through this multiplication, the denominator becomes a rational number. Specifically, the result of multiplying \((6+\sqrt{3})(6-\sqrt{3})\) results in \(6^2 - (\sqrt{3})^2 = 36 - 3 = 33\). See how this cleared the denominator of radicals? This kind of operation resembles the principle seen in the difference of squares, which states \((a-b)(a+b) = a^2 - b^2\). This method is key when simplifying expressions, making them easier to interpret and work with in future calculations.

Binomial Expansion and Its Role

When simplifying expressions involving radicals, binomial expansion is often used. Binomial expansion refers to expanding an expression raised to a power, especially expressions of the form \((a+b)(a-b)\) which results in difference of squares. Even though in our particular problem the expression \(6+\sqrt{3}\) was not directly expanded into numerous terms, understanding binomial expansion is pivotal when multiplying factors such as \((6+\sqrt{3})(6-\sqrt{3})\).

In expanding such a binomial, each part of one binomial is multiplied by each part of the other binomial. This precisely leads to what is known as the difference of squares formula where \(a^2 - b^2\) is the result. In our context, this made solving and simplifying the radical expression seamless by transforming it into a more manageable form. Mastering binomial expansion and recognizing these patterns simplify complex algebraic manipulations and are invaluable in learning mathematics holistically.