Stitz-Zeager_College_Algebra_e-book

2 sequences and the binomial theorem summation

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Unformatted text preview: E1 + E2 (E 1) x2 + y 2 = 4 (E 2) y 2 + y = 4 √ From y 2 + y = 4 we get y 2 + y − 4 = 0 which gives y = −1± 17 . Due to the complicated 2 nature of these answers, it is worth our time to make a quick sketch of both equations to head off any extraneous solutions we may encounter. We see that the circle x2 + y 2 = 4 intersects the parabola y = x2 exactly twice,√and both of these points have a positive y value. Of the two solutions for y , only y = −1+ 17 is positive, so to get our solution, we substitute this 2 1 We encourage the reader to solve the system using substitution to see that you get the same solution. 534 Systems of Equations and Matrices √ √ −1+ 17 2 √ = 0 and solve for x. We get x = ± = ± −2+2 17 . Our solution is 2 √ √ √ √ √ −2+2 17 −1+ 17 −2+2 17 −1+ 17 ,− , which we leave to the reader to verify. , , 2 2 2 2 into y − √ x2 y y 1 −3 −2 −1 1 1 2 3 −3 −2 x −1 1 2 3 x −1 Graphs for x2 + y 2 = 4 y − 2x = 0 Graphs for x2 + y 2 = 4 y − x2 = 36 A couple of remarks about Example 8.7.1 are in order. First note that, unlike systems of linear equations, it is possible for a system of non-linear equations to have more than one solution without having infinitely many solutio...
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