Algebra 2 Practice - Rationalizing the Denominator (Example 1)

Please subscribe!    / nickperich   *Algebra 2 Practice - Rationalizing the Denominator* Rationalizing the denominator is the process of eliminating any square roots (or other radicals) from the denominator of a fraction. This is typically done by multiplying both the numerator and denominator by a value that will make the denominator a rational number. --- *Step 1: Rationalizing a Single Square Root in the Denominator* When you have a square root in the denominator, multiply both the numerator and denominator by the same square root. This eliminates the square root from the denominator. #### *Example 1: Rationalizing the Denominator* Simplify the expression \( \frac{5}{\sqrt{2}} \). 1. *Multiply both the numerator and denominator by \( \sqrt{2} \):* \[ \frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \] Now the denominator is rationalized, and the expression is: \[ \frac{5\sqrt{2}}{2} \] --- *Step 2: Rationalizing a Binomial Denominator (Conjugates)* If the denominator is a binomial with a square root, you rationalize by multiplying both the numerator and denominator by the conjugate of the denominator. The *conjugate* of a binomial \( a + b\sqrt{c} \) is \( a - b\sqrt{c} \), and vice versa. #### *Example 2: Rationalizing a Binomial Denominator* Simplify the expression \( \frac{3}{1 + \sqrt{5}} \). 1. *Multiply both the numerator and denominator by the conjugate of \( 1 + \sqrt{5} \), which is \( 1 - \sqrt{5} \):* \[ \frac{3}{1 + \sqrt{5}} \times \frac{1 - \sqrt{5}}{1 - \sqrt{5}} = \frac{3(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})} \] 2. *Simplify the denominator using the difference of squares formula* \( (a + b)(a - b) = a^2 - b^2 \): \[ (1 + \sqrt{5})(1 - \sqrt{5}) = 1^2 - (\sqrt{5})^2 = 1 - 5 = -4 \] 3. **Simplify the numerator**: \[ 3(1 - \sqrt{5}) = 3 - 3\sqrt{5} \] Now the expression is: \[ \frac{3 - 3\sqrt{5}}{-4} = \frac{-3 + 3\sqrt{5}}{4} \] So, the rationalized expression is: \[ \frac{-3 + 3\sqrt{5}}{4} \] --- *Step 3: Rationalizing Denominators with Higher Roots* If the denominator involves a cube root or higher root, you may need to multiply both the numerator and denominator by a value that makes the denominator a perfect cube (or higher power) in order to eliminate the radical. #### *Example 3: Rationalizing a Cube Root Denominator* Simplify \( \frac{4}{\sqrt[3]{2}} \). 1. *Multiply both the numerator and denominator by \( \sqrt[3]{4} \), because \( 2 \times 4 = 8 \), and \( \sqrt[3]{8} = 2 \):* \[ \frac{4}{\sqrt[3]{2}} \times \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = \frac{4\sqrt[3]{4}}{\sqrt[3]{8}} \] 2. **Simplify the denominator**: \[ \sqrt[3]{8} = 2 \] So, the expression becomes: \[ \frac{4\sqrt[3]{4}}{2} \] 3. **Simplify the fraction**: \[ \frac{4\sqrt[3]{4}}{2} = 2\sqrt[3]{4} \] Thus, the rationalized expression is: \[ 2\sqrt[3]{4} \] --- **Summary**: **Rationalizing a square root**: Multiply both the numerator and denominator by the same square root to eliminate the square root in the denominator. **Rationalizing a binomial denominator**: Multiply both the numerator and denominator by the conjugate of the binomial. **Rationalizing higher roots**: Multiply both the numerator and denominator by a value that makes the denominator a perfect cube or higher power, depending on the root. By following these steps, you can eliminate square roots or other radicals from the denominator and make the expression simpler and more manageable. I have many informative videos for Pre-Algebra, Algebra 1, Algebra 2, Geometry, Pre-Calculus, and Calculus. Please check it out: / nickperich Nick Perich Norristown Area High School Norristown Area School District Norristown, Pa #math #algebra #algebra2 #maths #math #shorts #funny #help #onlineclasses #onlinelearning #online #study