Dividing Radicals Using Conjugates | MathHelp.com - By MathHelp.com
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| 00:0-1 | to simplify this problem , we must get rid of | |
| 00:03 | the radical in the denominator but notice that there are | |
| 00:07 | two terms in the denominator of this fraction . In | |
| 00:12 | this situation , in order to get rid of the | |
| 00:15 | radical , we must multiply top and bottom of the | |
| 00:18 | fraction by the conjugate of the denominator . The conjugate | |
| 00:24 | of two plus route seven is 2 -17 . So | |
| 00:30 | here we multiply top and bottom of the fraction by | |
| 00:34 | two route seven , Distributing in the numerator . We | |
| 00:42 | have 10 -5 . Route seven . In the denominator | |
| 00:50 | notice that we must foil two times 2 is four | |
| 01:00 | . two times negative route seven is -2 , Routes | |
| 01:05 | seven , Positive route seven times 2 is positive to | |
| 01:10 | route seven And positive route seven times negative . Route | |
| 01:14 | seven is negative . Route 49 or negative seven . | |
| 01:22 | Notice that in the denominator the minus two , Routes | |
| 01:25 | seven plus two . Route seven cancel each other out | |
| 01:30 | And we have 4 -7 which is -3 . So | |
| 01:36 | we have 10 -5 . Route seven over negative three | |
| 01:45 | . As your last step , I would change the | |
| 01:47 | minus sign in the denominator to A plus , which | |
| 01:52 | changes the sign of every term in the numerator . | |
| 01:56 | So we have -10-plus 5 , 7 over positive three | |
| 00:0-1 | . |
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