Algebra Notes Quadratic Systems Name: Block: Date: Last class we discussed linear systems. The only possibilities we had we 1 solution, no solution or infinite solutions. With quadratic systems we have a few more options. No Solution (graphs do not intersect) One Solution Two Solutions Three Solutions Four Solutions Remember: The solution to the system is the set of POINTS that satisfy the system! SUBSTITUTION: The first equation is a quadratic equation and the second is a linear equation. Notice that we have a single y variable in each equation. If we solve the second equation for y and use substitution, we should be able to find the solution. y 6 Now we can solve the system 6 0 3 4 0 ( 4)( 1) 4, 1 y Since both equations equal y we can set them equal Combine like terms Factor Solve for Now that we know the value of we can substitute each one back into the linear equation to find the y values. y (4) 6 This gives of the points (4,6) and (- 1,-4) as solutions to our system. It is important that we check our solutions y ( 1) 4 using BOTH equations. We can also represent our solution graphically. If we graph both y 6 equations we see that y they do indeed intersect at (-1,-4) and (4,6).
Eample : Solve the following system y = + (y ) = 4 From the form of the equations, you should know that this system contains a parabola and a circle (remember those from geometry?). According to the graph, there should be three solutions to this system: Using the first equation, I think I'll substitute "y" for " " in the second equation, and solve: + (y ) = 4 y + (y ) = 4 y + (y 4y + 4) = 4 y 3y = 0 y(y 3) = 0 y = 0, y = 3 Now I need to find the corresponding -values. When y = 0: y = 0 = 0 = (This is the solution at the origin that we'd been epecting.) When y = 3: y = 3 = 3 Then the solutions are the points, (0, 0), and.
Substitution Eample 3 Solve the following system: y = + 3 + y = + 3 I'll set the equations equal, and solve. I see that I cannot factor my equation so I use the Quadratic Formula. 3 3 1 0 Using the Quadratic Formula I get Plugging in back into the linear equation gives us the corresponding y value. Plugging in back into the linear equation gives us the corresponding y value. Our solutions are
Substitution Eample 4: y 3 Solve the following system: y 17 Graphically, this system is a straight line crossing a circle centered at the origin: There appear to be two solutions. I'll proceed algebraically to confirm this impression, and to get the eact values. Since the first equation is already solved for y, I will plug " 3" in for "y" in the second equation, and solve for the values of : + y = 17 + ( 3) = 17 + ( 3)( 3) = 17 + ( + 6 + 9) = 17 + 6 + 9 = 17 + 6 8 = 0 + 3 4 = 0 ( + 4)( 1) = 0 = 4, = 1 When = 4, y = 3 = ( 4) 3 = 4 3 = 1 When = 1, y = 3 = (1) 3 = 4 Then the solution consists of the points ( 4, 1) and (1, 4). Note the procedure: I solved one of the equations (the first equation looked easier) for one of the variables (solving for "y=" looked easier), and then plugged the resulting epression back into the other equation. This gave me one equation in one variable (the variable happened to be ), and a one-variable equation is something I know how to solve. Once I had the solution values for, I back-solved for the corresponding y-values. I emphasize "corresponding" because you have to keep track of which y-value goes with which -value. In the eample above, the points ( 4, 4) and (1, 1) are not solutions. Even though I came up with = 4 and 1 and y = 4 and 1, the = 4 did not go with the y = 4, and the = 1 did not go with the y = 1. Warning: You must match the -values and y-values correctly!
ELIMINATION (COMBINATION): You only want to use this method if you can eliminate both the squared term and the linear term for one of the variables. Consider y y 8 0 7 0 y y 9 8 4 7 0 If I multiply the first equation by 1 and add it to the second equation then I can eliminate both the variables. y y 8 0 7 0 y y 9 8 4 7 0 8y 4y 0 I can then solve by factoring, completing the square or using the quadratic formula. I choose factoring! 8y 4y 0 8 yy ( 3) 0 y 0, 3 Now I will substitute my y values back into one of my original equations and solve for y. Let s start with 0. y y 8 0 7 0 (0) 8 0(0) 7 0 8 7 0 ( 1)( 7) 0 So my solutions for when y 0 are 1,0 and 7,0 1, 7 When y 3 we get y y 8 0 7 0 ( 3) 8 0( 3) 7 0 9 8 60 7 0 8 76 0 I can t factor formula. 8 76 0, so I ll use the quadratic 8 8 4(1)(76) 8 40 (1) Since I m taking the square root of a negative, there are no real solutions when y 3. Thus, my only solutions are 1,0 and 7,0