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 theEquation 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 intoThe 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 – 16x + 39 = 0
• 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