Problem 59 - Algebra for College Students

Question

Rationalize the denominator and simplify. All variables represent positive real numbers. \(\frac{\sqrt{2}}{\sqrt{10}-\sqrt{3}}\)

Step-by-Step Solution

Answer

The expression simplifies to \(\frac{2\sqrt{5} + \sqrt{6}}{7}\).

Step 1: Identify the Need to Rationalize the Denominator

The expression \(\frac{\sqrt{2}}{\sqrt{10}-\sqrt{3}}\) has a denominator in the form of \(\sqrt{a} - \sqrt{b}\). To rationalize it, we need to eliminate the square roots in the denominator by multiplying by its conjugate, \(\sqrt{10} + \sqrt{3}\).

Step 2: Set Up the Conjugate

The conjugate of the denominator \(\sqrt{10} - \sqrt{3}\) is \(\sqrt{10} + \sqrt{3}\). We will multiply both the numerator and the denominator by this conjugate.

Step 3: Multiply by the Conjugate

Multiply the numerator and denominator by \(\sqrt{10} + \sqrt{3}\) to get: \(\frac{\sqrt{2} \cdot (\sqrt{10} + \sqrt{3})}{(\sqrt{10} - \sqrt{3})(\sqrt{10} + \sqrt{3})}\).

Step 4: Simplify the Numerator

Distribute \(\sqrt{2}\) over \(\sqrt{10} + \sqrt{3}\):\(\sqrt{2} \cdot \sqrt{10} = \sqrt{20},\) \(\sqrt{2} \cdot \sqrt{3} = \sqrt{6}\).Thus, the numerator is \(\sqrt{20} + \sqrt{6}\).

Step 5: Simplify the Denominator

Apply the difference of squares formula: \((a - b)(a + b) = a^2 - b^2\).For \(a = \sqrt{10}\) and \(b = \sqrt{3}\):\[\text{Denominator} = (\sqrt{10})^2 - (\sqrt{3})^2 = 10 - 3 = 7\].

Step 6: Complete the Rationalization

Substitute the simplified numerator and denominator back into the expression:\[ \frac{\sqrt{20} + \sqrt{6}}{7} \].

Step 7: Simplify Further if Necessary

Simplify \(\sqrt{20}\): \(\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}\).Hence, the expression can be further simplified to:\[ \frac{2\sqrt{5} + \sqrt{6}}{7} \].

Key Concepts

Simplifying Radicals

Simplifying radicals involves expressing a square root in its simplest form. This process helps make equations easier to work with and understand.
Consider the number under the square root, known as the radicand. The aim is to break down the radicand into perfect squares. A perfect square is a number like 4, 9, or 16, because they can be square rooted into integers:

• For \(\sqrt{20}\), notice that 20 can be expressed as \(4 \times 5\).
• Since 4 is a perfect square, it can be simplified: \(\sqrt{4} = 2\).
• Thus, \(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\).

Breaking down radicals into their simplest form involves identifying and simplifying all possible perfect squares within the radicand, ensuring equations remain neat and manageable.

Difference of Squares Formula

The difference of squares formula is a handy tool in algebra, especially when rationalizing expressions. Famously written as \(a^2 - b^2 = (a-b)(a+b)\), it allows us to transform complex expressions into simpler components.
This formula is particularly useful when dealing with expressions that involve a sum and difference of the same term, as seen here:

• In the problem, the denominator \(\sqrt{10} - \sqrt{3}\) and its conjugate \(\sqrt{10} + \sqrt{3}\) form a difference of squares: \((\sqrt{10})^2 - (\sqrt{3})^2\).
• Apply the formula to achieve a simple result: \(10 - 3 = 7\).

This reduces sqrt-based expressions efficiently and is vital when trying to eliminate radicals from the denominator.

Conjugates in Algebra

Conjugates simplify expressions, particularly when rationalizing denominators. They come into play when the denominator consists of radicals.
A conjugate reverses the sign between two terms while keeping the rest of the expression intact. For example, the conjugate of \(\sqrt{10} - \sqrt{3}\) is \(\sqrt{10} + \sqrt{3}\). When multiplying by a conjugate:

• The radicals in the denominator disappear, making the expression real and easier to handle.
• As in our example, multiplying by the conjugate leads to the application of the difference of squares formula, yielding a non-radical result.

By using conjugates, expressions become simpler and more refined, which is especially helpful in algebraic operations and solving equations involving radicals.