Solving Radical Equations | MathHelp.com - By MathHelp.com
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| 00:0-1 | we start this problem by isolating the radical on the | |
| 00:03 | left side of the equation . So subtracting six from | |
| 00:08 | both sides , We have the squared of x equals | |
| 00:15 | X -6 . Next to get rid of the radical | |
| 00:20 | . We square both sides of the equation And notice | |
| 00:27 | that on the right side of the equation the X | |
| 00:30 | -6 squared must be thought of as X -6 times | |
| 00:35 | X -6 . So we have X equals x minus | |
| 00:41 | six times X minus six . And foiling the right | |
| 00:47 | side , we have X equals x squared minus 12 | |
| 00:53 | . X plus 36 . Remember that when an equation | |
| 00:59 | has a squared term in it , We must set | |
| 01:02 | the equation equal to zero then factor . So our | |
| 01:07 | next step is to move the X to the right | |
| 01:10 | side by subtracting X from both sides of the equation | |
| 01:16 | to get zero equals x squared minus 13 X plus | |
| 01:22 | 36 . Which then factors as zero equals x minus | |
| 01:30 | nine times X minus four , solving from here , | |
| 01:38 | X equals nine or X equals four . And as | |
| 01:45 | your last step , make sure to check both answers | |
| 01:49 | back in the original equation . If we plug a | |
| 01:53 | nine in for X , we have the squared of | |
| 01:56 | nine or three plus six equals nine . And since | |
| 02:03 | three plus six equals nine is a true statement , | |
| 02:06 | our first answer works , but notice that when we | |
| 02:09 | plug a four back in for X in the original | |
| 02:12 | equation we have the squared of four , which is | |
| 02:15 | two plus six equals four and two plus six does | |
| 02:22 | not equal four , so four cannot be a solution | |
| 02:26 | And our answer is only x equals nine . |
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