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Chapter 16 Quadratic Equations

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**Solving Quadratic Equations by Completing the Square**

§ 16.2 Solving Quadratic Equations by Completing the Square

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Completing the Square Solving a Quadratic Equation by Completing a Square If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. Isolate all variable terms on one side of the equation. Complete the square (half the coefficient of the x term squared, added to both sides of the equation). Factor the resulting trinomial. Use the square root property.

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**Solving Equations Example Solve by completing the square. y2 + 6y = 8**

y = 4 or 2

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**Solving Equations Example (y + ½)2 = Solve by completing the square.**

y2 + y – 7 = 0 y2 + y = 7 y2 + y + ¼ = 7 + ¼ (y + ½)2 =

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**Solving Equations Example Solve by completing the square.**

2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½ x2 + 7x = ½ = (x + )2 =

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**Solving Quadratic Equations by the Quadratic Formula**

§ 16.3 Solving Quadratic Equations by the Quadratic Formula

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The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.

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The Quadratic Formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.

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**The Quadratic Formula Example**

Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1

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**The Quadratic Formula Example**

Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = (multiply both sides by 8) a = 1, b = 8, c = 20

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**The Quadratic Formula Example**

Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.

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**Graphs of Quadratic Equations**

Example x y Graph y = 2x2 – 4. (–2, 4) (2, 4) x y 2 4 1 –2 (–1, – 2) (1, –2) –4 –1 –2 (0, –4) –2 4

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**Intercepts of the Parabola**

Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y.

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**Characteristics of the Parabola**

If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. 2) the x-coordinate of the vertex is To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.

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**Graphs of Quadratic Equations**

Example Graph y = –2x2 + 4x + 5. x y (1, 7) Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex is (0, 5) (2, 5) x y 3 –1 (–1, –1) (3, –1) 2 5 1 7 5 –1 –1

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