+2446 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 \((0,1)\cup (4,+\infty)\)
.+2446 It appears as if this question has been edited since the replies of other answerers. As of now, the expression to simplify is \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
| \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) | Radicals in the denominator are considered to be improper; it is required that all radical terms are moved to the numerator. This can be achieved by utilizing the conjugate of the denominator: \(2-\sqrt{3}\). |
| \(\frac{2-\sqrt{3}}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}\) | To simplify the denominator, it is possible to use the algebraic rule that \((a+b)(a-b)=a^2-b^2\). Notice that this process does not change the inherent value of the expression since all I am doing is multiplying by 1, which has no effect. |
| \(\frac{\left(2-\sqrt{3}\right)^2}{2^2-\sqrt{3}^2}\) | By squaring the radical in the denominator, we are getting rid of a radical's existence from the denominator. |
| \(\frac{\left(2-\sqrt{3}\right)^2}{4-3}\) | Well, it is very clear at this point that the denominator simplifies to 1. Expand the numerator using the rule that \((a-b)^2=a^2-2ab+b^2\) |
| \(2^2-2*2*\sqrt{3}+\sqrt{3}^2\) | Simplify. |
| \(4-4\sqrt{3}+3\) | |
| \(7-4\sqrt{3}\) | This is what OfficialBubbleTanks got. |
+2446 In the rational function \(f(x)=\frac{2x}{x^2-5x-14}\), we can determine the vertical asymptotes by setting the denominator equal to 0 and solving for x.
| \(x^2-5x-14=0\) | In this case, the left-hand-side quadratic is factorable, which eases the process of finding the zeros. |
| \((x-7)(x+2)=0\) | Use the zero product thereom to find the remaining zeros. |
| \(x_1=a=7\\ x_2=b=-2\) | |
The horizontal asymptote requires some observation. The horizontal asymptote lies on y=0 since the degree of the numerator is less than the degree of the denominator. Therefore, c=0. We know the unknown variables, so we can now calculate their collective sum.
\(a+b+c\\ 7-2+0\\ 5\)
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