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