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 $8t^2 \le 3-10t$.

Guest Feb 27, 2018

#1**+1 **

1. Let's just pretend for the moment that \(c^3+4c>5c^2\) is an equation, so the equation would be \(c^3+4c=5c^2\). Let's solve this as if it were a regular equation.

\(c^3+4c=5c^2\) | Subtract 5c^2 from both sides so that every term containing a "c" is on the left. |

\(c^3-5c^2+4c=0\) | Factor out the GCF of the left-hand-side of the equation, c. |

\(c(c^2-5c+4)=0\) | 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." |

\(c_1=0\\ c_2=1\\ c_3=4\) | |

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:

\(c<0\\ 0 < c < 1\\ 0 < c < 4\\ c>4\)

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.

\(c^3+4c>5c^2\) | 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. |

\((-1)^3+4*-1\ngtr5(-1)^2\) | 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.

\(c^3+4c>5c^2\) | Plug in the desired point. |

\(\left(\frac{1}{2}\right)^3+4*\frac{1}{2}>5*\left(\frac{1}{2}\right)^2\) | This one does not appear to be as clear-cut, so we will have to perform more simplification. |

\(\frac{1}{8}+2>\frac{5}{4}\) | 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.

\(c^3+4c>5c^2\) | Evaluate if true when c=2. |

\(2^3+4*2>5*2^2\) | 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. |

\(8+8\ngtr20\) | As aforementioned, \(1 is not in the range. |

And finally, \(c>4\) is the last one. Let's do the next integer, c=5.

\(5^3+5*2>5^1*5^2\) | You can use any tricks you desire to evaluate these inequalities quicker. Here is the one that I noticed straightaway. |

\(5^3+5*2=5^3\) | 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 \((0,1)\cup (4,+\infty)\)

TheXSquaredFactor
Feb 27, 2018

#1**+1 **

Best Answer

1. Let's just pretend for the moment that \(c^3+4c>5c^2\) is an equation, so the equation would be \(c^3+4c=5c^2\). Let's solve this as if it were a regular equation.

\(c^3+4c=5c^2\) | Subtract 5c^2 from both sides so that every term containing a "c" is on the left. |

\(c^3-5c^2+4c=0\) | Factor out the GCF of the left-hand-side of the equation, c. |

\(c(c^2-5c+4)=0\) | 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." |

\(c_1=0\\ c_2=1\\ c_3=4\) | |

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:

\(c<0\\ 0 < c < 1\\ 0 < c < 4\\ c>4\)

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.

\(c^3+4c>5c^2\) | 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. |

\((-1)^3+4*-1\ngtr5(-1)^2\) | 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.

\(c^3+4c>5c^2\) | Plug in the desired point. |

\(\left(\frac{1}{2}\right)^3+4*\frac{1}{2}>5*\left(\frac{1}{2}\right)^2\) | This one does not appear to be as clear-cut, so we will have to perform more simplification. |

\(\frac{1}{8}+2>\frac{5}{4}\) | 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.

\(c^3+4c>5c^2\) | Evaluate if true when c=2. |

\(2^3+4*2>5*2^2\) | 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. |

\(8+8\ngtr20\) | As aforementioned, \(1 is not in the range. |

And finally, \(c>4\) is the last one. Let's do the next integer, c=5.

\(5^3+5*2>5^1*5^2\) | You can use any tricks you desire to evaluate these inequalities quicker. Here is the one that I noticed straightaway. |

\(5^3+5*2=5^3\) | 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 \((0,1)\cup (4,+\infty)\)

TheXSquaredFactor
Feb 27, 2018