Question

Need a quick run through on 'rationalizing denominators'. Any takers?

Answer 1

depends on denominator

Answer 2

\[\frac{3}{\sqrt{2}}\] use
\[\frac{3}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\] to get
\[\frac{3\sqrt{2}}{2}\]

Answer 3

for example... \[6/3-\sqrt{13}\]

Answer 4

or if
\[\frac{x}{x-\sqrt{3}}\] use
\[\frac{x}{x-\sqrt{3}}\times \frac{x+\sqrt{3}}{x+\sqrt{3}}\] to get
\[\frac{x(x+\sqrt{3})}{x^2-3}\]

Answer 5

ah like second example

Answer 6

1 possibility:
\[\frac{something}{\sqrt{a}+\sqrt{b}}*\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}\]
another:
\[\frac{something}{\sqrt{a}-\sqrt{b}}*\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}\]
another:
\[\frac{something}{a-\sqrt{b}}*\frac{a+\sqrt{b}}{a+\sqrt{b}}\]
another:
\[\frac{something}{a+\sqrt{b}}*\frac{a-\sqrt{b}}{a-\sqrt{b}}\]
anohter:
\[\frac{something}{\sqrt{a}}*\frac{\sqrt{a}}{\sqrt{a}}\]
and the list continues....

Answer 7

in your example multiply top and bottom by "conjugate" of bottom

Answer 8

\[\frac{6}{3-\sqrt{13}}\times \frac{3+\sqrt{13}}{3+\sqrt{13}}\]

Answer 9

the conjugate of a+b is a-b

Answer 10

the conjugate of a-b is a+b

Answer 11

how would you rationalize the denominator of satellite73's example?

Answer 12

denominator is
\[9-14=-5\] numerator is what you bet via distributive law

Answer 13

how did you get -5 from the denominator :
\[3-\sqrt{13}\]

Answer 14

\[(3-\sqrt{13})(3+\sqrt{13})=3*3+3\sqrt{13}-3\sqrt{13}-\sqrt{13}\sqrt{13}=9+0-13=-4\]

Answer 15

he made a mistake

Answer 16

this is confusing