Problem 41 - Intermediate Algebra

Question

Rationalize each denominator. See Example 4. $$ \frac{6}{2-\sqrt{7}} $$

Step-by-Step Solution

Answer

The rationalized form is \(-4 - 2\sqrt{7}\).

Step 1: Identify the Denominator

Notice that the denominator of the fraction is \(2 - \sqrt{7}\). We need to rationalize this denominator by removing the square root.

Step 2: Find the Conjugate

To remove the square root from the denominator, multiply the numerator and the denominator by the conjugate of \(2 - \sqrt{7}\), which is \(2 + \sqrt{7}\).

Step 3: Multiply by the Conjugate

Multiply both the numerator and the denominator by the conjugate: \[\frac{6}{2 - \sqrt{7}} \times \frac{2 + \sqrt{7}}{2 + \sqrt{7}}\] This gives us: \[\frac{6(2 + \sqrt{7})}{(2 - \sqrt{7})(2 + \sqrt{7})}\]

Step 4: Simplify the Numerator

Distribute \(6\) into \(2 + \sqrt{7}\): \[6 \times 2 + 6 \times \sqrt{7} = 12 + 6\sqrt{7}\] So the new numerator is \(12 + 6\sqrt{7}\).

Step 5: Simplify the Denominator

Use the difference of squares formula to simplify the denominator:\[(2 - \sqrt{7})(2 + \sqrt{7}) = 2^2 - (\sqrt{7})^2 = 4 - 7 = -3\] So, the denominator simplifies to \(-3\).

Step 6: Final Simplification

Simplify the whole expression:\[\frac{12 + 6\sqrt{7}}{-3} = \frac{12}{-3} + \frac{6\sqrt{7}}{-3} = -4 - 2\sqrt{7}\]Therefore, the rationalized form is \(-4 - 2\sqrt{7}\).

Key Concepts

Conjugates in Algebra

When dealing with algebraic expressions that contain square roots, like in the equation \( \frac{6}{2-\sqrt{7}} \), we often encounter the concept of conjugates. Conjugates are used to remove the square root from the denominator or numerator of fractions, helping to simplify the expression.

The conjugate of any binomial expression, \(a + b\), is \(a - b\) and vice versa. In our original problem, the conjugate of \(2 - \sqrt{7}\) is \(2 + \sqrt{7}\). By multiplying both the numerator and the denominator of the fraction by this conjugate, we can eliminate the square root from the denominator.

• When multiplying two conjugates, use the formula: \((a+b)(a-b) = a^2 - b^2\).
• For example: \((2 - \sqrt{7})(2 + \sqrt{7})\) results in \(2^2 - (\sqrt{7})^2 = 4 - 7 = -3.\)

This step drastically changes the expression, making it easier to handle without square roots.

Difference of Squares

The difference of squares is a powerful algebraic property used for simplifying expressions with conjugates. It is based on the identity \((a+b)(a-b) = a^2 - b^2\), where \(a\) and \(b\) can be numbers or expressions. This technique is crucial when rationalizing denominators that involve square roots.

For our example, the denominator \((2 - \sqrt{7})(2 + \sqrt{7})\) simplifies using this property. You get \(2^2\) which is 4, and \((\sqrt{7})^2\) which is 7. When you subtract these, the result is \(4 - 7 = -3.\)

• This subtraction process effectively removes the square root, leaving a simple number.
• It clears up the fraction, preventing any irrational number in the denominator.

Using this formula not only simplifies the algebraic work but also aligns with standard math practices for clear solutions.

Simplifying Radicals

Simplifying radicals means making a mathematical expression with roots, such as square roots, easier to read and work with. When you multiply radicals together, using properties like conjugates and the difference of squares helps to clear up complex terms.

In our solution, we've simplified the terms by multiplying by conjugates and applied the difference of squares. This simplifies the original equation \(\frac{6}{2-\sqrt{7}}\) to \(-4 - 2\sqrt{7}\). Here's how to understand simplifying radicals effectively:

• Multiply radicals with their conjugates to remove square roots from denominators.
• Use formulas like the difference of squares \((a^2 - b^2)\) for faster computation.
• Follow up by simplifying each term, like \(\frac{12 + 6\sqrt{7}}{-3}\) breaking down to \(-4 - 2\sqrt{7}.\)

This explanation highlights not just the process, but the logic behind simplifying complex radical expressions into manageable forms.