University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

45

Pages:

3
Author:

ignorantkd77o

Practice with Radicals Core Concepts Product Property of Square Roots Words: The square root of a product equals the product of the square roots of the factors. Numbers: [\sqrt{9 \cdot 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}] or [\sqrt{6} \cdot \sqrt{3} = \sqrt{6 \cdot 3} = \sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}] Algebra: [\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}, \text{ where } a,b \geq 0] Quotient Property of Square Roots Words: The square root of a quotient equals the quotient of the square roots of the numerator and denominator. Numbers: [\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}] Algebra: [\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \text{ where } a \geq 0 \text{ and } b > 0] IF [\frac{4}{\sqrt{3}}] Piece of Advice Treat √ like variables [\frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}] multiplying by the √ will get rid of the root [\frac{2}{\sqrt{5}} + 4\sqrt{5} = 4\sqrt{5} + \frac{2}{\sqrt{5}}] [= \frac{20}{5} + 4\sqrt{5}] [= 4 + 4\sqrt{5}] 9.1 Extra Practice In Exercises 1-12, simplify the expression. [\sqrt{24} = \sqrt{4 \cdot 6} = \sqrt{4} \cdot \sqrt{6} = 2\sqrt{6}] [\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}] [\sqrt{162g^6} = \sqrt{81 \cdot 2 \cdot g^6} = 9g^3\sqrt{2}] [-\sqrt{512h^7} = -\sqrt{256 \cdot 2 \cdot h^6 \cdot h} = -16h^3\sqrt{2h}] [\sqrt{\frac{25}{64}} = \frac{\sqrt{25}}{\sqrt{64}} = \frac{5}{8}] [-\sqrt{\frac{6}{49}} = -\frac{\sqrt{6}}{\sqrt{49}} = -\frac{\sqrt{6}}{7}] [\sqrt{\frac{196}{r^4}} = \frac{\sqrt{196}}{\sqrt{r^4}} = \frac{14}{r^2}] [\sqrt{\frac{49x^3}{64y^2}} = \frac{\sqrt{49x^3}}{\sqrt{64y^2}} = \frac{7x\sqrt{x}}{8y}] [\sqrt[3]{-135} = -3\sqrt[3]{5}] [\sqrt[3]{729} = 9] [-\sqrt[3]{192x^5} = -4x\sqrt[3]{3x^2}] [\sqrt[3]{\frac{12a^6}{512b^4}} = \frac{\sqrt[3]{12a^6}}{\sqrt[3]{512b^4}} = \frac{a^2\sqrt[3]{12}}{8b}]