Rationalizing the Denominator
Algebra 1 - Unit 3, Day 4
Today's Objective
Do Now: Radical Multiplication
Simplify each expression.
\( \sqrt{5} \cdot \sqrt{5} = ? \)
\( \sqrt{3} \cdot \sqrt{6} = ? \)
\( (2\sqrt{7}) \cdot \sqrt{7} = ? \)
How to Rationalize the Denominator
The rule is simple: You are not allowed to leave a radical in the denominator. Here's how we fix it, step-by-step:
Identify the square root in the denominator.
Multiply both the top and the bottom of the fraction by that same square root.
Simplify. The denominator should now be a rational number (no more root!).
Practice: Rationalizing
I Do
\( \frac{2}{\sqrt{5}} \)
\( \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \)
\( \frac{2\sqrt{5}}{5} \)
We Do
\( \frac{1}{\sqrt{3}} \)
\( \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \)
\( \frac{\sqrt{3}}{3} \)
You Do
\( \frac{7}{\sqrt{2}} \)
\( \frac{7}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \)
\( \frac{7\sqrt{2}}{2} \)
I Do: \( \frac{2}{\sqrt{5}} \) = \( \frac{2\sqrt{5}}{5} \)
We Do: \( \frac{1}{\sqrt{3}} \) =
You Do: \( \frac{7}{\sqrt{2}} \) =
Practice: Simplify, then Rationalize
I Do
\( \frac{5}{\sqrt{12}} \)
\( \frac{5}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \)
\( \frac{5\sqrt{3}}{2 \cdot 3} \)
\( \frac{5\sqrt{3}}{6} \)
We Do
\( \frac{6}{\sqrt{8}} \)
\( \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}} \)
\( \frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \)
\( \frac{3\sqrt{2}}{2} \)
You Do
\( \sqrt{\frac{7}{20}} \)
\( \frac{\sqrt{7}}{\sqrt{20}} = \frac{\sqrt{7}}{2\sqrt{5}} \)
\( \frac{\sqrt{7}}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \)
\( \frac{\sqrt{35}}{10} \)
I Do: \( \frac{5}{\sqrt{12}} \) = \( \frac{5\sqrt{3}}{6} \)
We Do: \( \frac{6}{\sqrt{8}} \) =
You Do: \( \sqrt{\frac{7}{20}} \) =
Independent Practice
Simplify each expression completely.
Level 1 Problems:
1. \( \frac{1}{\sqrt{7}} \)
2. \( \frac{10}{\sqrt{5}} \)
Level 2 Problems:
3. \( \frac{2}{\sqrt{18}} \)
4. \( \sqrt{\frac{3}{5}} \)
Level 3 Problem:
5. \( \frac{3\sqrt{2}}{\sqrt{6}} \)
Independent Practice
1. \( \frac{1}{\sqrt{7}} \) =
2. \( \frac{10}{\sqrt{5}} \) =
3. \( \frac{2}{\sqrt{18}} \) =
4. \( \sqrt{\frac{3}{5}} \) =
5. \( \frac{3\sqrt{2}}{\sqrt{6}} \) =
Exit Ticket
Simplify the expression completely:
\( \frac{7}{\sqrt{28}} \)
\( \frac{7}{\sqrt{4 \cdot 7}} = \frac{7}{2\sqrt{7}} \)
\( \frac{7}{2\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{7\sqrt{7}}{2 \cdot 7} = \frac{7\sqrt{7}}{14} \)
Answer: \( \frac{\sqrt{7}}{2} \)