+0

0
52
1

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
+1807
+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
edited by TheXSquaredFactor  Feb 27, 2018
Sort:

#1
+1807
+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
edited by TheXSquaredFactor  Feb 27, 2018

### 10 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details