The question asks about the type of roots--not what the roots are.
Because of this technicality, we can combine knowledge from Descartes' Rule of Signs and the Fundamental Theorem of Algebra and the Complex Conjugate Root Theorem in order to answer this question.
The Descartes' Rule of Signs states that the number of sign changes in \(f(x)\), when the terms are ordered by degree, indicates the number of positive solutions or less than this by an even number. Let's consider the original function \(f(x)=5x^2-7x-6\):
\(f(x)=\underbrace{5x^2-7}\underbrace{x-6}\\ \hspace{16mm}+1\hspace{7mm}+0\hspace{10mm}=\text{1 sign change}\)
Notice that +5x2 and -7x have a different sign, so I marked that change with a +1. -7x and -6 do not have a signage change, so I marked it with a +0.
The Fundamental Theorem of Algebra is relevant because it indicates how many total solutions that a function can have. In this case, there are 2 roots since the degree of the polynomial is 2.
The Complex Conjugate Root Theorem states that if polynomials have real coefficients with complex roots, then they will come in pairs; in other words, there will always be an even number of non-real roots.
I would say that it would be wise to consolidate this information into a table.
Number of Positive Solutions | Number of Negative Solutions | Number of Complex Roots | Total Number of Roots | |
1 | 1 | 0 | 2 | |
Based on the rules mentioned previously, only one possibility remains: Both roots are real.
I think this question is one where some people might want to pull out their hair in frustration. I suggest gathering all your information into an organized table.
Number of Rooms Painted | Rate | Time | ||
Paulo | ||||
Irina | ||||
Paulo + Irina | ||||
We may not be able to fill in all the information, but that is OK! Do not forget what the end goal is! The end goal is to figure out how much time it will take if Paulo and Irina work together. This is the information we know:
Now that we know all this information, let's fill in the table.
Number of Rooms Painted | Rate | Time | ||
Paulo | 1 | 8 hours | ||
Irina | 1 | 9 hours | ||
Paulo + Irina | 1 | |||
Now, what is the painting speed of Paulo? I think you would agree that Paulo can paint at 1 room per 8 hours. This is the rate for Paulo! We can use the same logic for Irina. Irina paints at a speed of 1 room per 9 hours. Let's fill that information in!
Number of Rooms Painted | Rate | Time | ||
Paulo | 1 | \(\frac{1\text{ room}}{8\text{ hours}}\) | 8 hours | |
Irina | 1 | \(\frac{1\text{ room}}{9\text{ hours}}\) | 9 hours | |
Paulo + Irina | 1 | |||
In order to fill in the rest of the table, we will have to introduce variables! I will use "x" to represent the number of hours it takes for Paulo and Irina collectively to paint this room. This means that the rate of them together is \(\frac{1\text{ room}}{x \text{ hours}}\). Let's fill that in, as well!
Number of Rooms Painted | Rate | Time | ||
Paulo | 1 | \(\frac{1\text{ room}}{8\text{ hours}}\) | 8 hours | |
Irina | 1 | \(\frac{1\text{ room}}{9\text{ hours}}\) | 9 hours | |
Paulo + Irina | 1 | \(\frac{1\text{ room}}{x\text{ hours}}\) | x hours | |
We can assume that, if Paulo and Irina work together at their individual rates, then the rate will be the sum.
Therefore, \(\frac{1}{8}+\frac{1}{9}=\frac{1}{x}\). Let's solve this!
\(\frac{1}{8}+\frac{1}{9}=\frac{1}{x}\) | The best way, I believe, to solve this equation is to multiply by the LCM of the denominators. in this case, that number would be \(72x\). |
\(9x+8x=72\) | Add the like terms. |
\(17x=72\) | Divide by 17 on both sides. |
\(x=\frac{72}{17}\approx\text{4.24 hours}\) | |
Therefore, the correct answer choice is the last one (which is really just a manipulation of the equation I created).
a)
If we suppose that c varies inversely with d, then this is what that statement is telling you:
The most basic example of this occurence is \(y=\frac{1}{x}\). As you think of larger values of x, the input, then the output, y, will decrease.
There is only one difference from the previous function to this particular problem: We do not know the rate in which the inverse relation occurs. For example, if the numerator of the previous function was a 2, then the rate of the inverse function would double. Let's name the rate of the inverse relation "k." We, therefore, make the following equation.
\(c=\frac{k}{d}\)
b)
We can use the previous equation to figure out the rest. It involves basic substitution.
\(c=\frac{k}{d}\) | Substitute in the known values for c and d. |
\(17=\frac{k}{2}\) | Solve for k by multiplying by 2 on both sides. |
\(k=34\) | |
Now, let's find d when c=68:
\(c=\frac{k}{d}\) | We now know the values of c and k are this time. |
\(68=\frac{34}{d}\) | Now it is a matter of simplifying. |
\(68d=34\) | |
\(d=\frac{34}{68}=\frac{1}{2}=0.5\) | |