528 Review Questions and Answers

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When we subtract the quantity (a 2x 2 + b 2) from each side, we get 2abx = 2abx We can add 2abx to each side and then divide through by 4ab (which is legal because we ve been told that a 0 and b 0, so we can be sure that 4ab 0) When we do that, we obtain x = 0 as the only root of the mixed quadratic We can substitute this value for x into the original equations, obtaining y = b2 in both cases Therefore, the system has the single solution (0,b2)

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Question 27-8

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Imagine that we re trying to solve a general two-by-two system of equations, and we mix them to get a single equation in one variable When we solve that mixed equation, we get two different roots, one of which has multiplicity 1, and the other of which has multiplicity 2 What does this say about the solutions of the original system

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Answer 27-8

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The solutions of the original two-by-two system have the same multiplicity pattern as the roots of the mixed equation In this case, that means there is one solution with multiplicity 1, and another solution with multiplicity 2

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Question 27-9

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Consider the following pair of equations: y = (x + 1)3 and y = x 3 + 2x 2 + x How can we find the real solutions to this two-by-two system

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Answer 27-9

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Let s multiply out the first equation to get it into polynomial standard form When we do that, the system becomes y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 2x 2 + x We can mix the right sides of these equations, getting x 3 + 3x 2 + 3x + 1 = x 3 + 2x 2 + x

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Part Three 529

Now let s subtract the entire right side of this equation from the left side When we do that, we obtain x 2 + 2x + 1 = 0 which factors into (x + 1)2 = 0 This equation has the single real root x = 1, with multiplicity 2 When we plug that into the first original equation, we get y = (x + 1)3 = ( 1 + 1)3 = 03 =0 Therefore, the original system has the single real solution ( 1,0), with multiplicity 2

Question 27-10

Consider the following pair of equations: y = (x + 1)3 and y = (x + 2)3 How can we find the real solutions of this two-by-two system, if any exist

Answer 27-10

Let s multiply both of these equations out to get cubics in polynomial standard form That gives us y = x 3 + 3x 2 + 3x + 1 and y = x 3 + 6x 2 + 12x + 8 When we mix the right sides of these equations, we get x 3 + 3x 2 + 3x + 1 = x 3 + 6x 2 + 12x + 8 Subtracting the entire left side from the right side and then switching right-to-left, we obtain 3x 2 + 9x + 7 = 0

530 Review Questions and Answers

The discriminant of this quadratic is negative, telling us that it has no real roots The x-values of any solutions we can derive for the original system will not be real numbers Therefore, no real solutions exist

28

Question 28-1

How do we graph a general two-by-two system of equations when we want to see approximately where the curves intersect at the real solutions, but we don t need a lot of precision

Answer 28-1

First, we can calculate several ordered pairs for both functions individually, including the real solutions, if any exist It can be helpful to put the values in a table Next, we figure out the scales we should have on our graph so as to provide a good picture of the situation Then we plot the real solution point or points, if any exist, on the coordinate grid After that, we plot the rest of the points based on the values in the table we ve created Finally, we fill in the lines or curves for both functions

Question 28-2

Consider the system of equations we solved in Answer 27-2: y = x2 and y = x 2 How can we sketch an approximate graph of this system, showing the real solution

Answer 28-2

We can tabulate and plot several points in both functions including the real solution, (0,0) Table 30-1 compares some values of x, some values of the first function, and some values of

Table 30-1

Selected values for graphing the functions y = x 2 and y = x 2 The bold entry indicates the real solution