## Presentation on theme: "Solving Quadratics by Completing the Square & Quadratic Formula"— Presentation transcript:

By: Jeffrey Bivin Lake Zurich High School Last Updated: October 24, 2007

X2 + 6x + 9 x 1 1 1 1 1 1 x x + 3 Now, complete the square 1 + 9 1 1
Jeff Bivin -- LZHS

X2 + 4x + 4 4 x 1 1 1 1 x x + 2 Now, complete the square 1 1 x + 2
Jeff Bivin -- LZHS

X2 + 5x + 25/4 x 1 1 .5 1 1 1 x x + 5/2 Now, complete the square 1
Jeff Bivin -- LZHS

X2 - 6x + 9 x 1 1 1 1 1 1 1 1 1 x - 3 x 1 + 9 1 1 turn 1 square over
turn 2 squares over turn 2 squares over turn 3 squares over Jeff Bivin -- LZHS

Solve by Completing the Square
x2 + 10x + 8 = 0 (x2 + 10x ) = -8 (x2 + 10x + (5)2) = (5)2 = 25 (x + 5)2 = 17 Jeff Bivin -- LZHS

Solve by Completing the Square
3x2 + 24x + 12 = 0 3 x2 + 8x + 4 = 0 (x2 + 8x ) = -4 (x2 + 8x + (4)2) = (4)2 = 16 (x + 4)2 = 12 Jeff Bivin -- LZHS

Solve by Completing the Square
2x2 + 5x - 12 = 0 2 Jeff Bivin -- LZHS

Solve by Completing the Square
2x2 - 12x - 11 = 0 2 Jeff Bivin -- LZHS

Solve by Completing the Square
-5x2 + 12x + 19 = 0 -5 Jeff Bivin -- LZHS

Solve by Completing the Square
5x2 - 30x + 45 = 0 5 Jeff Bivin -- LZHS

Solve by Completing the Square
5x2 - 30x + 75 = 0 5 Jeff Bivin -- LZHS

Convert to vertex form y = x2 + 10x + 8 y - 8 = (x2 + 10x )
(5)2 = 25 y = (x2 + 10x + (5)2) y + 17 = (x2 + 10x + (5)2) y + 17 = (x + 5)2 - 17 x + 5 = 0 Axis of symmetry: x = -5 Vertex: (-5, -17) Jeff Bivin -- LZHS

Convert to vertex form y = 5x2 - 30x + 46 y - 46 = 5(x2 - 6x )
5(-3)2 = 45 y = 5(x2 - 6x + (-3)2) y - 1 = 5(x2 - 6x + (-3)2) y - 1 = 5(x - 3)2 + 1 x - 3 = 0 Axis of symmetry: x = 3 Vertex: (3, 1) Jeff Bivin -- LZHS

Solve by Completing the Square
ax2 + bx + c = 0 a The Quadratic Formula Jeff Bivin -- LZHS

3x2 + 7x - 4 = 0 a = 3 b = 7 c = -4 Jeff Bivin -- LZHS

6x2 + 9x + 2 = 0 a = 6 b = 9 c = 2 Jeff Bivin -- LZHS

5x2 - 8x + 1 = 0 a = 5 b = -8 c = 1 Jeff Bivin -- LZHS

6x2 - 17x - 14 = 0 a = 6 b = -17 c = -14 Jeff Bivin -- LZHS

x2 + 6x + 9 = 0 a = 1 b = 6 c = 9 Jeff Bivin -- LZHS

3x2 + 7x + 5 = 0 a = 3 b = 7 c = 5 Two Imaginary Solutions Jeff Bivin -- LZHS

Why do some quadratic equations have 2 real solutions some have 1 real solution and some have two imaginary solutions?

Now Consider Discriminant ax2 + bx + c = 0
89 -11 If discriminant > 0, then 2 real solutions If discriminant = 0, then 1 real solution If discriminant < 0, then 2 imaginary solutions Jeff Bivin -- LZHS

That's All Folks Jeff Bivin -- LZHS