10.7 Solving Quadratic Systems p. 632

We’ve already studied two techniques for solving systems of linear equations. You will use these same techniques to solve quadratic systems. These techniques are ??? Substitution Linear combination

Find the points of intersection x2 + y2 = 13 & y = x + 1 We will use….. Substitution. x2 + (x + 1)2 = 13 x2 + (x2 + 2x + 1) = 13 2x2 + 2x – 12 = 0 2(x2 + x – 6) = 0 2(x + 3)(x – 2) = 0 x = -3 & x = 2 Now plug these values into the Equation to get y!! (-3,-2) and (2,3) are the points where the two graphs intersect. Check it on your calculator!

Your turn! Find the points of intersection of: x2 + y2 = 5 & y = -x + 3 (1,2) & (2,1)

Solve by substitution: x2 + 4y2 – 4 = 0 -2y2 + x + 2 = 0 The second equation has no x2 term so solve for x → x = 2y2 – 2 and substitute it into the first equation. (2y2 – 2)2 + 4y2 - 4 = 0 4y4 – 8y2 + 4 + 4y2 – 4 = 0 4y4 – 4y2 = 0 4y2 (y2 – 1) = 0 4y2 (y-1)(y+1) = 0 y = 0, y = 1, y = -1 Now plug these x values into The revised equation Which gives you : (-2,0) (0,1) (0,-1)

Linear combination x2 + y2 – 16x + 39 = 0 x2 – y2 – 9 = 0 If you add these two equations together, the y’s will cancel x2 – y2 - 9 = 0

x2 + y2 – 16x + 39 = 0 x2 – y2 - 9 = 0 2x2 – 16x + 30 = 0 2(x2 – 8x + 15) = 0 2 (x-3) ( x-5) = 0 x = 3 or x = 5 Plugging these into one of the ORIGINAL equations to get: (3,0) (5,4) ( 5,-4)

Assignment