We’ll have a test on polynomials and rational functions (corresponding to the material in Chapter 3 of the textbook, but also including the material on graphing functions found in Chapter 2) on Sunday, November 2nd.
Note that calculators will not be permitted for this test.
The following questions will help you to review.
Page 87 question 12
Page 89 question 23
Pages 149–150 questions 4, 5, 7, 11, 14, 17, 22, 28
Markus asked a good question about question 4 on page 149. One way to solve this question (the hard way!) is to find 3 equations in 3 unknowns. However, there’s an easier way… in fact, only the information in i) and ii) is needed to answer the question (and no calculator is required!).
Hint: Think about the various forms of a quadratic expression.
Dear Dr. McDonald,
I was able to solve question 4 without a calculator. However I can’t solve question 5 at all and 7 only with a calculator.
Is there another way to solve question 7 without a calculator? And can you please give me any hints for question 5.
Thank you,
Markus
Hi Markus,
I should note that—while calculators will not be used on the test—questions from the textbook may require a calculator. (It’s useful to be able to distinguish those questions that do require a calculator from those that don’t.) In general, if a question requires that you solve a system of 3 equations in 3 unknowns (as, I think, is the case in question 7), you can expect that a calculator will be allowed.
However, question 5 can be completed without a calculator, and you will probably find the Sum and Product Theorem useful.
Dear Dr. McDonald!
I am not sure how to answer question 11 properly. In the textbook the solution just says k is element of R.
I used the discrimant and since there have to be 2 real solutions the discriminant hast to be greater than 0. So i calculated the discriminant and got 8k^2 -20k+24>0
How shall i proceed now?
Good work so far!
To summarise what you’ve got so far, you want to find the values of \(k\) that will make the discriminant positive (hence yielding two distinct real solutions for \(x\)), and the work you’ve done to this point shows that you’ll get two distinct real solutions for \(x\) whenever the value you get for \(k\) satisfies \[8k^2-20k+24>0\]
Which values of \(k\) will make this true? Well, the expression involving \(k\) is just a quadratic (in the variable \(k\)), so its graph will be a parabola, and it will be concave up since the lead coefficient is 8. What information does the discriminant of this quadratic expression provide for you?
The discriminant is a complex number, therefore the parabola will not cut the x-axis, so y can not be negative or 0. That means k is an element of R. Is that right?
You’ve got the right idea, just be careful about how you express your result. The discriminant is a negative number, and so the parabola never crosses the x axis. That means it is either always positive, or always negative. Since the parabola is concave up, the only way this can happen (not crossing the axis) is if the parabola is completely above the x axis.
So, that means that, for any value of \(k\) (so, any \(k \in \mathbb{R}\)), the expression involved in the inequality is always greater than 0. Thus, any value of \(k\) will yield two unique (real) solutions for \(x\).
Ok thank you Sir, I get it now!