Math Problem Statement

Rationalize the denominator

Solution

Rationalizing the denominator means eliminating any irrational numbers, like square roots, from the denominator of a fraction. We usually achieve this by multiplying both the numerator and the denominator by a term that will make the denominator rational.

Here's a general approach based on the type of denominator:

1. Denominator with a Single Square Root

If the denominator has a single square root, like in ab\frac{a}{\sqrt{b}}:

  • Multiply both the numerator and the denominator by b\sqrt{b} to get: ab×bb=abb\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a \sqrt{b}}{b}

2. Denominator with a Binomial Involving Square Roots

If the denominator is a binomial with a square root, like ab+c\frac{a}{b + \sqrt{c}}:

  • Multiply both the numerator and the denominator by the conjugate of the denominator, which is bcb - \sqrt{c}: ab+c×bcbc=a(bc)b2c\frac{a}{b + \sqrt{c}} \times \frac{b - \sqrt{c}}{b - \sqrt{c}} = \frac{a(b - \sqrt{c})}{b^2 - c}
  • This results in a rational denominator because the difference of squares is applied.

Would you like a specific example solved, or do you have a particular problem in mind?

Follow-up Questions:

  1. How do you rationalize a denominator with multiple square roots?
  2. What is the conjugate, and why is it useful in rationalization?
  3. How does rationalization change if cube roots are involved?
  4. How would you rationalize a denominator with both square roots and variables?
  5. Why is rationalizing the denominator often required in mathematical expressions?

Tip: Conjugates are especially helpful when dealing with binomials, as they allow simplification through the difference of squares formula.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rationalization
Radicals
Square Roots

Formulas

For \( \frac{a}{\sqrt{b}} \), multiply by \( \frac{\sqrt{b}}{\sqrt{b}} \): \( \frac{a \sqrt{b}}{b} \)
For \( \frac{a}{b + \sqrt{c}} \), multiply by conjugate \( \frac{b - \sqrt{c}}{b - \sqrt{c}} \): \( \frac{a(b - \sqrt{c})}{b^2 - c} \)

Theorems

Difference of Squares Theorem

Suitable Grade Level

Grades 9-11