Problem 72

Question

Rationalize the denominator and simplify. All variables represent positive real numbers. \(\frac{\sqrt{x}+1}{\sqrt{x}-10}\)

Step-by-Step Solution

Verified
Answer
The rationalized expression is \(\frac{x + 11\sqrt{x} + 10}{x - 100}\).
1Step 1: Identify the Conjugate of the Denominator
The expression involves a fraction with a square root in the denominator. To rationalize, we need to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{x} - 10\) is \(\sqrt{x} + 10\).
2Step 2: Multiply Numerator and Denominator by the Conjugate
Multiply the entire fraction by \(\frac{\sqrt{x} + 10}{\sqrt{x} + 10}\), which equals 1, to ensure we are only modifying the expression, not its value. This yields:\[\frac{(\sqrt{x} + 1)(\sqrt{x} + 10)}{(\sqrt{x} - 10)(\sqrt{x} + 10)}\]
3Step 3: Expand the Numerator
Distribute the terms in the numerator:\[(\sqrt{x} + 1)(\sqrt{x} + 10) = \sqrt{x} \cdot \sqrt{x} + \sqrt{x} \cdot 10 + 1 \cdot \sqrt{x} + 1 \cdot 10 = x + 10\sqrt{x} + \sqrt{x} + 10\]Combine the like terms:\[x + 11\sqrt{x} + 10\]
4Step 4: Simplify the Denominator
Apply the formula for the difference of squares: \((a - b)(a + b) = a^2 - b^2\). Here, \(a = \sqrt{x}\) and \(b = 10\), simplify as follows:\[\sqrt{x}^2 - 10^2 = x - 100\]
5Step 5: Write the Final Simplified Expression
Combine your results from steps 3 and 4 to get the final expression:\[\frac{x + 11\sqrt{x} + 10}{x - 100}\]
6Step 6: Verify the Combination Validity
Ensure all terms are simplified and correctly combined. There are no like terms across the numerator and denominator to further simplify.

Key Concepts

ConjugateDifference of SquaresSimplifying Expressions
Conjugate
When you encounter a square root in the denominator of a fraction, a helpful trick is to use the concept of a conjugate to eliminate it. The conjugate of a binomial expression, like \(a - b\) or \(a + b\), is another expression with the same terms but the opposite sign between them. For instance, the conjugate of \(a - b\) is \(a + b\).

Using the conjugate involves multiplying the fraction by a special form of 1, \(\frac{a+b}{a+b}\) or \(-\frac{a-b}{a-b}\), so you don't change the fraction's value. This method simplifies the process of dealing with square roots in denominators.

Here’s why it works: when you multiply by the conjugate, you create a pattern that neatly eliminates the square root from the denominator without affecting the original value of the expression.
Difference of Squares
The difference of squares is an algebraic identity that simplifies expressions of the form \(a^2 - b^2\). It works because it involves two terms that are perfect squares subtracted from each other. The formula is:
  • \((a - b)(a + b) = a^2 - b^2\)
When you multiply a binomial by its conjugate, you apply this identity, resulting in the elimination of the square root. This is especially useful in rationalizing denominators.

For example, multiplying \(\sqrt{x} - 10\) by \(\sqrt{x} + 10\) produces:
  • \((\sqrt{x})^2 - 10^2 = x - 100\)
This simplification turns the denominator into an expression without radicals, making it much easier to handle numerically and algebraically.
Simplifying Expressions
Simplifying expressions is all about combining like terms and eliminating unnecessary complexity to make mathematical expressions easier to work with. This can involve several different algebraic operations, such as distribution, factoring, and reducing fractions.

In the context of rationalizing a denominator, like in our exercise, simplifying means taking the resulting expanded terms and combining them wherever possible. For instance, after expanding the numerator, you'll often need to do some cleanup by adding similar terms together.
  • Terms like \(10\sqrt{x} + \sqrt{x}\) become \(11\sqrt{x}\).
  • A final check ensures that no further simplification is possible, such as reducing fractions or canceling common factors.
This step ensures that we present the cleanest, most understandable form of the expression, ready for further use in calculations or problem solving.