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Solving Quadratics by Completing the Square & Quadratic Formula By: Jeffrey Bivin Lake Zurich High School Last Updated: October 24, 2007

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X 2 + 6x + 9 Jeff Bivin -- LZHS

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X 2 + 4x + 4 Jeff Bivin -- LZHS 4

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X 2 + 5x + 25/4 Jeff Bivin -- LZHS

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X 2 - 6x + 9 Jeff Bivin -- LZHS

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

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

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

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

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

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

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

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

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

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Solve by Completing the Square ax 2 + bx + c = 0 a Jeff Bivin -- LZHS

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Solve using the Quadratic Formula 3x 2 + 7x - 4 = 0 a = 3 b = 7 c = -4 Jeff Bivin -- LZHS

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Solve using the Quadratic Formula 6x 2 + 9x + 2 = 0 a = 6 b = 9 c = 2 Jeff Bivin -- LZHS

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Solve using the Quadratic Formula 5x 2 - 8x + 1 = 0 a = 5 b = -8 c = 1 Jeff Bivin -- LZHS

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Solve using the Quadratic Formula 6x x - 14 = 0 a = 6 b = -17 c = -14 Jeff Bivin -- LZHS

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Solve using the Quadratic Formula x 2 + 6x + 9 = 0 a = 1 b = 6 c = 9 Jeff Bivin -- LZHS

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Solve using the Quadratic Formula 3x 2 + 7x + 5 = 0 a = 3 b = 7 c = 5 Jeff Bivin -- LZHS

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

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

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Jeff Bivin -- LZHS

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