Polynomial Inequalities | MathHelp.com - By MathHelp.com
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| 00:0-1 | to solve a polynomial inequality , like the one shown | |
| 00:03 | here . Our first step is to write the corresponding | |
| 00:07 | equation . In other words , we simply change the | |
| 00:11 | inequality sign to an equal sign and we have x | |
| 00:16 | squared minus three equals nine minus x . Next we | |
| 00:22 | solve the equation since we have a squared term , | |
| 00:26 | We first set the equation equal to zero , So | |
| 00:30 | we moved the 9 -1 to the left side by | |
| 00:33 | subtracting nine and adding X to both sides of the | |
| 00:37 | equation . This gives us x squared plus x minus | |
| 00:43 | 12 equals zero . Next we factor the left side | |
| 00:48 | as the product of two by no meals . Since | |
| 00:51 | the factors of negative 12 that add to positive one | |
| 00:55 | are positive four and negative three , we have X | |
| 01:00 | plus four times X minus three equals zero , so | |
| 01:05 | either X plus four equals zero or x minus three | |
| 01:09 | equals zero . And solving each equation from here we | |
| 01:13 | have X equals negative four , or X equals three | |
| 01:19 | . Now it's important to understand that the solutions to | |
| 01:22 | the equation negative four and 3 represent what are called | |
| 01:27 | the critical values of the inequality and we plot these | |
| 01:32 | critical values on a number line . However , notice | |
| 01:36 | that our original inequality uses a greater than sign rather | |
| 01:41 | than a greater than or equal to sign . So | |
| 01:45 | we use open dots on our critical values of -4 | |
| 01:50 | and positive three . Remember that greater than or less | |
| 01:54 | than means an open dot and greater than or equal | |
| 01:58 | to or less than or equal to means a closed | |
| 02:01 | dot . Now we can see that are critical values | |
| 02:06 | have divided the number line into three separate intervals Less | |
| 02:11 | than -4 Between -4 and three And greater than three | |
| 02:17 | . And here's the important part . Our next step | |
| 02:20 | is to test a value from each of the intervals | |
| 02:24 | by plugging the value back into the original inequality to | |
| 02:28 | see if it gives us a true statement . So | |
| 02:31 | let's first test of value from the less than -4 | |
| 02:36 | interval Such as -5 . If we plug a negative | |
| 02:41 | five back in for both , X is in the | |
| 02:44 | original inequality we have negative five squared minus three is | |
| 02:50 | greater than nine minus a negative five , which simplifies | |
| 02:55 | to 25 minus three is greater than nine plus five | |
| 02:59 | , or 22 is greater than 14 , Since 22 | |
| 03:04 | is greater than 14 is a true statement . This | |
| 03:08 | means that all values in the interval . We're testing | |
| 03:11 | our solutions to the inequality so we shade the interval | |
| 03:17 | . Next we test a value from the between negative | |
| 03:20 | four and three interval , such as zero . If | |
| 03:24 | we plug a zero back in for both , X | |
| 03:27 | is in the original inequality we have zero squared minus | |
| 03:32 | three is greater than 9 0 , which simplifies to | |
| 03:38 | zero minus three is greater than nine or negative three | |
| 03:42 | is greater than nine , Since -3 is greater than | |
| 03:46 | nine is a false statement . This means that all | |
| 03:50 | values in the interval we're testing are not solutions to | |
| 03:54 | the inequality , so we don't shade the interval . | |
| 03:59 | Next we test a value from the greater than three | |
| 04:02 | interval , such as four . If we plug a | |
| 04:05 | four back in for both , X is in the | |
| 04:08 | original inequality we have four squared minus three is greater | |
| 04:13 | than nine minus four , which simplifies to 16 -3 | |
| 04:19 | is greater than five or 13 is greater than five | |
| 04:24 | , Since 13 is greater than five is a true | |
| 04:27 | statement . This means that all values in the interval | |
| 04:31 | , we're testing our solutions to the inequality so we | |
| 04:35 | shade the interval . Finally , we write the answer | |
| 04:39 | that's shown on our graph in set notation , The | |
| 04:43 | set of all X is such that X is less | |
| 04:46 | than -4 or X is greater than three . |
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