Please answer in interval notation
1. Find all values of $c$ such that $c^3 + 4c > 5c^2$. Answer with interval notation.
2. Find all values of $t$ such that $8t^2 \le 3-10t$.
+2446 1. Let's just pretend for the moment that c3+4c>5c2 is an equation, so the equation would be c3+4c=5c2. Let's solve this as if it were a regular equation.
| c3+4c=5c2 | Subtract 5c^2 from both sides so that every term containing a "c" is on the left. |
| c3−5c2+4c=0 | Factor out the GCF of the left-hand-side of the equation, c. |
| c(c2−5c+4)=0 | Now, let's factor some more. The trinomial is factorable. |
| c(c−4)(c−1)=0 | Now, let's solve for all the possible values of "c." |
| c1=0c2=1c3=4 | |
Therefore, we have defined many possible ranges where the solutions of this inequality can lie. We will have to test all of them. The ranges are as follows:
c<00<c<10<c<4c>4
Now, let's test values in these ranges.
For the range c<0, let's test a point in this range, say -1, and see if it satisfies the original equation.
| c3+4c>5c2 | Plug in -1 as a test point and see if it satisfies. |
| (−1)3+4∗−1>5(−1)2 | If you use some logic here, you will realize that this test point does not satisfy the original inequality. (-1)^3 and 4*-1 are both negative values, so adding them together will result in a negative value, too. On the right hand side, we have a number squared, which will always be positive. Multiplying this result by 5 does not affect the signage. Therefore, the right hand side cannot be greater than the left, so it is a false statement. |
| (−1)3+4∗−1≯5(−1)2 | This means that any numbers in the range c<0 are not solutions to the original inequality. |
Now, let's try the range \(0 . I will choose 1/2.
| c3+4c>5c2 | Plug in the desired point. |
| (12)3+4∗12>5∗(12)2 | This one does not appear to be as clear-cut, so we will have to perform more simplification. |
| 18+2>54 | Here, you can stop. 2 is already greater than 5/4. |
| This means that any real number in the range \(0 is a solution to the original inequality. |
Now, let's try the terciary range of \(1 . My test point, in this case, will be 2. Always pick a point that is easiest to plug in. Overcomplicating matters is only a burden to yourself.
| c3+4c>5c2 | Evaluate if true when c=2. |
| 23+4∗2>5∗22 | This one appears to require some more iterations to determine. |
| 8+8>20 | I think it is obvious at this point that this is untrue. |
| 8+8≯20 | As aforementioned, \(1 is not in the range. |
And finally, c>4 is the last one. Let's do the next integer, c=5.
| 53+5∗2>51∗52 | You can use any tricks you desire to evaluate these inequalities quicker. Here is the one that I noticed straightaway. |
| 53+5∗2=53 | Well, there is a 5^3 on both sides of the inequality. On the left, a 5*2 remains, so the left side has to larger. Notice how I refrained from actually evaluating this completely as to preserve precious time. |
| c>4 is the range as well. |
Consolidating all this information, we can conclude that \(0 or c>4. When you convert to inequality notation, you get (0,1)∪(4,+∞)
+2446 1. Let's just pretend for the moment that c3+4c>5c2 is an equation, so the equation would be c3+4c=5c2. Let's solve this as if it were a regular equation.
| c3+4c=5c2 | Subtract 5c^2 from both sides so that every term containing a "c" is on the left. |
| c3−5c2+4c=0 | Factor out the GCF of the left-hand-side of the equation, c. |
| c(c2−5c+4)=0 | Now, let's factor some more. The trinomial is factorable. |
| c(c−4)(c−1)=0 | Now, let's solve for all the possible values of "c." |
| c1=0c2=1c3=4 | |
Therefore, we have defined many possible ranges where the solutions of this inequality can lie. We will have to test all of them. The ranges are as follows:
c<00<c<10<c<4c>4
Now, let's test values in these ranges.
For the range c<0, let's test a point in this range, say -1, and see if it satisfies the original equation.
| c3+4c>5c2 | Plug in -1 as a test point and see if it satisfies. |
| (−1)3+4∗−1>5(−1)2 | If you use some logic here, you will realize that this test point does not satisfy the original inequality. (-1)^3 and 4*-1 are both negative values, so adding them together will result in a negative value, too. On the right hand side, we have a number squared, which will always be positive. Multiplying this result by 5 does not affect the signage. Therefore, the right hand side cannot be greater than the left, so it is a false statement. |
| (−1)3+4∗−1≯5(−1)2 | This means that any numbers in the range c<0 are not solutions to the original inequality. |
Now, let's try the range \(0 . I will choose 1/2.
| c3+4c>5c2 | Plug in the desired point. |
| (12)3+4∗12>5∗(12)2 | This one does not appear to be as clear-cut, so we will have to perform more simplification. |
| 18+2>54 | Here, you can stop. 2 is already greater than 5/4. |
| This means that any real number in the range \(0 is a solution to the original inequality. |
Now, let's try the terciary range of \(1 . My test point, in this case, will be 2. Always pick a point that is easiest to plug in. Overcomplicating matters is only a burden to yourself.
| c3+4c>5c2 | Evaluate if true when c=2. |
| 23+4∗2>5∗22 | This one appears to require some more iterations to determine. |
| 8+8>20 | I think it is obvious at this point that this is untrue. |
| 8+8≯20 | As aforementioned, \(1 is not in the range. |
And finally, c>4 is the last one. Let's do the next integer, c=5.
| 53+5∗2>51∗52 | You can use any tricks you desire to evaluate these inequalities quicker. Here is the one that I noticed straightaway. |
| 53+5∗2=53 | Well, there is a 5^3 on both sides of the inequality. On the left, a 5*2 remains, so the left side has to larger. Notice how I refrained from actually evaluating this completely as to preserve precious time. |
| c>4 is the range as well. |
Consolidating all this information, we can conclude that \(0 or c>4. When you convert to inequality notation, you get (0,1)∪(4,+∞)
