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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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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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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
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