Problem 58 - Algebra for College Students

Question

Rationalize the denominator and simplify. All variables represent positive real numbers. \(\frac{3}{\sqrt{3}+\sqrt{10}}\)

Step-by-Step Solution

Answer

The simplified form is \(-\frac{3}{7}(\sqrt{3} - \sqrt{10})\).

Step 1: Identify the Conjugate

To rationalize the denominator of the expression \( \frac{3}{\sqrt{3}+\sqrt{10}} \), we need to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \( \sqrt{3}+\sqrt{10} \) is \( \sqrt{3}-\sqrt{10} \).

Step 2: Multiply by the Conjugate

Multiply the numerator and the denominator of the expression by \( \sqrt{3}-\sqrt{10} \): \[\frac{3}{\sqrt{3}+\sqrt{10}} \cdot \frac{\sqrt{3}-\sqrt{10}}{\sqrt{3}-\sqrt{10}} = \frac{3(\sqrt{3}-\sqrt{10})}{(\sqrt{3}+\sqrt{10})(\sqrt{3}-\sqrt{10})}.\]

Step 3: Simplify the Denominator

Simplify the expression in the denominator using the difference of squares formula: \((a+b)(a-b) = a^2-b^2\). Here, \(a=\sqrt{3}\) and \(b=\sqrt{10}\). So, \[(\sqrt{3})^2 - (\sqrt{10})^2 = 3 - 10 = -7.\] The denominator is now \(-7\).

Step 4: Simplify the Numerator

Distribute the \(3\) in the numerator: \[3(\sqrt{3}) - 3(\sqrt{10}) = 3\sqrt{3} - 3\sqrt{10}.\]

Step 5: Final Expression

Combine the simplified numerator and denominator: \[\frac{3\sqrt{3} - 3\sqrt{10}}{-7} = \frac{-1}{7}(3\sqrt{3} - 3\sqrt{10}).\] Simplify by factoring out \(-3\) from the numerator:\[-\frac{3}{7}(\sqrt{3} - \sqrt{10}).\]

Key Concepts

Conjugate of a Binomial

When you encounter a binomial in the denominator, such as \( \sqrt{3} + \sqrt{10} \), you can use its conjugate to simplify expressions. But what is a conjugate? Simply put, the conjugate of a binomial \( a + b \) is \( a - b \).

The conjugate is useful because multiplying a binomial by its conjugate eliminates the irrational part of the expression. It effectively leverages the difference of squares, a fundamental principle in algebra, to convert a difficult radical expression into something simpler.

In our example, the conjugate of \( \sqrt{3} + \sqrt{10} \) is \( \sqrt{3} - \sqrt{10} \). By multiplying both the numerator and denominator by this conjugate, we can rationalize the denominator.

Difference of Squares

The concept of the difference of squares is a powerful algebraic tool that comes into play when rationalizing denominators.
The difference of squares formula is given by:

• \( (a+b)(a-b) = a^2 - b^2 \)

This formula is extremely handy because it lets us transform products of binomials into differences of two perfect squares.

Applying this to \( (\sqrt{3} + \sqrt{10})(\sqrt{3} - \sqrt{10}) \), we use \( a = \sqrt{3} \) and \( b = \sqrt{10} \). Hence, the multiplication simplifies to \( \sqrt{3}^2 - \sqrt{10}^2 = 3 - 10 = -7 \).

The resulting value, \(-7\), is a rational number, thus eliminating the radicals from the denominator.

Simplification of Radical Expressions

Simplifying radical expressions is essential for making equations easier to work with and understand. When we started with the expression \( \frac{3(\sqrt{3} - \sqrt{10})}{-7} \), our goal was to make it as simple as possible for further calculations.

First, we distribute the 3 in the numerator, producing \( 3\sqrt{3} - 3\sqrt{10} \). Then, combine this with the denominator to achieve the fraction form. Since the numerator can be factored, we extract \(-3\) out of the expression:

• The original expression becomes \(-\frac{3}{7}(\sqrt{3} - \sqrt{10})\).

This entire process allows students to deal with cleaner and rational expressions, making subsequent calculations straightforward.