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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
+2117
+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
#1
+2117
+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