Week 4: Multiplying and Dividing Radical Expressions

Published by on February 25, 2018 | Leave a response

*Simplifying Expressions with Numerical Radicands:
-Example: (Mulitiplying coeficients, radicands seperately; then add up the like terms)
1. (4\sqrt{2} + 2\sqrt{5})(3\sqrt{2} - \sqrt{5})
=4\sqrt{2}.3\sqrt{2} - 4\sqrt{2}.\sqrt{5} + 2\sqrt{5}.3\sqrt{2} - 2\sqrt{5}.\sqrt{5}
=12\sqrt{2.2}-4\sqrt{2.5} + 6\sqrt{5.2} - 2\sqrt{5.5}
=12\sqrt{4} - 4\sqrt{10} + 6\sqrt{10} - 2\sqrt{25}
=12.2 - 4\sqrt{10} + 6\sqrt{10} - 2.5
=(24 - 10) + (6\sqrt{10} - 4\sqrt{10})
=14+2\sqrt{10}
2. (5\sqrt{3} + 3\sqrt{7})(5\sqrt{3} - 3\sqrt{7})
=5\sqrt{3}.5\sqrt{3} - 5\sqrt{3}.3\sqrt{7} + 3\sqrt{7}.5\sqrt{3} - 3\sqrt{7}.3\sqrt{7}
=25\sqrt{3.3} - 9\sqrt{7.7}
=25.3 – 9.7
=12
3. (\sqrt{10}-\sqrt{7})
=(\sqrt{10}-\sqrt{7})(\sqrt{10}-\sqrt{7})
=\sqrt{10}.\sqrt{10} - \sqrt{10}.\sqrt{7} - \sqrt{7}.\sqrt{10} + \sqrt{7}.\sqrt{7}
=\sqrt{100} - \sqrt{70} - \sqrt{70} +\sqrt{49}
=10 + 7 - (\sqrt{70} + \sqrt{70})
=17 - 2\sqrt{70}
4. (7+2\sqrt{a})(3 - 5\sqrt{a}) (a \geq 0)
=7.3 - 35\sqrt{a} + 6\sqrt{a} - 10\sqrt{a.a}
=21 - 29\sqrt{a} - 10a
*Dividing: Express as fractions – Rationalizing
** A Monomial Denominator: Multiply the numerator and the denominator by the denominator without coefficient.
-Example:
1. \frac{3\sqrt{7} + 5}{\sqrt{7}}
=\frac{3\sqrt{7} + 5}{\sqrt{7}} . \frac{\sqrt{7}}{\sqrt{7}}
=\frac{3\sqrt{7}.\sqrt{7} + 5.\sqrt{7}}{\sqrt{7}.\sqrt{7}}
=\frac{3\sqrt{49} + 5\sqrt{7}}{\sqrt{49}}
=\frac{3.7 + 5\sqrt{7}}{7}
=\frac{21 + 5\sqrt{7}}{7}
2. \frac{6\sqrt{a} - 4\sqrt{b}}{2\sqrt{ab}} (a>0 and b>0)
=\frac{6\sqrt{a} - 4\sqrt{b}}{2\sqrt{ab}} . \frac{\sqrt{ab}}{\sqrt{ab}}
=\frac{6\sqrt{aab} - 4\sqrt{abb}}{2\sqrt{aabb}}
=\frac{6a\sqrt{b} - 4b\sqrt{a}}{2ab}
=\frac{3a\sqrt{b} - 2b\sqrt{a}}{ab}
** A Binomial Denominator: Multiply the numerator and the denominator by the conjugate of the denominator
-Example:
1. \frac{4 + 2\sqrt{6}}{3\sqrt{2} - 4}
=\frac{4 + 2\sqrt{6}}{3\sqrt{2} - 4} . \frac{3\sqrt{2} + 4}{3\sqrt{2} + 4}
=\frac{4(3\sqrt{2} + 4) + 2\sqrt{6}(3\sqrt{2} + 4)}{3\sqrt{2}(3\sqrt{2} + 4) - 4(3\sqrt{2} + 4)}
=\frac{12\sqrt{2} + 16 + 6\sqrt{12} + 8\sqrt{6}}{9\sqrt{4} + 12\sqrt{2} - 12\sqrt{2} - 16}
=\frac{12\sqrt{2} + 16 + 6\sqrt{4.3} + 8\sqrt{6}}{9.4 - 16}
=\frac{12\sqrt{2} + 16 + 12\sqrt{3} + 8\sqrt{6}}{20}
=\frac{3\sqrt{2} + 4 + 3\sqrt{3} + 2\sqrt{6}}{5}
2. \frac{2\sqrt{a} - 3\sqrt{b}}{2\sqrt{a} + 3\sqrt{b}} (a>0 and b>0)
=\frac{2\sqrt{a} - 3\sqrt{b}}{2\sqrt{a} + 3\sqrt{b}}. \frac{2\sqrt{a} - 3\sqrt{b}}{2\sqrt{a} - 3\sqrt{b}}
=\frac{2\sqrt{a}.(2\sqrt{a} - 3\sqrt{b}) - 3\sqrt{b}.(2\sqrt{a} - 3\sqrt{b})}{2\sqrt{a}(2\sqrt{a} - 3\sqrt{b}) + 3\sqrt{b}(2\sqrt{a} - 3\sqrt{b})}
=\frac{4\sqrt{aa} - 6\sqrt{ab} - 6\sqrt{ab} + 9\sqrt{bb}}{4\sqrt{aa} - 6\sqrt{ab} + 6\sqrt{ab} - 9\sqrt{b}}
=\frac{4a - 12\sqrt{ab} + 9b}{4a - 9b}
***Optional: Some regular formulars can be memorized to reduce the process
1. a^{2} - b^{2} = (a-b)(a+b)
2. (a+b)^{2} = a^{2} +2ab +b^{2}
3. (a-b)^{2} = a^{2} -2ab +b^{2}

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