# Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 16 Quadratic Equations.

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Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 16 Quadratic Equations

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 16.1 Solving Quadratic Equations by the Square Root Property

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Square Root Property We previously have used factoring and the zero- factor property to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If x 2 = a for a ≥ 0, then

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: x 2 = 49 Solve: (y – 3) 2 = 4 Solve: 2x 2 = 4 x 2 = 2 y = 3 ± 2 y = 1 or 5 Example Square Root Property

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: x 2 + 4 = 0 x 2 = –4 There is no real solution because the square root of –4 is not a real number. Square Root Property Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: (x + 2) 2 = 25 x = –2 ± 5 x = –2 + 5 or x = –2 – 5 x = 3 or x = –7 Square Root Property Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: (3x – 17) 2 = 28 3x – 17 = Square Root Property Example

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 16.2 Solving Quadratic Equations by Completing the Square

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. The three quadratic expressions above have something in common. They are all perfect square trinomials. Notice that in each case, the constant term is the square of half the coefficient of the x term in the trinomial. (This is true as long as the coefficient of the x 2 term is 1). Completing the Square x 2 + 18 + 81 x 2 – 12x + 36 x 2 – 10x + 25

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. What constant term should be added to the following expressions to create a perfect square trinomial? x 2 – 10x add 5 2 = 25 x 2 + 16x add 8 2 = 64 x 2 – 7x add Completing the Square Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Completing the Square Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 12 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve a Quadratic Equation by Completing a Square Step 1:If the coefficient of x 2 is 1, go to Step 2. If not, divide both sides of the equation by the coefficient of x 2. Step 2:Get all terms with variables on one side of the equation and constants on the other side. Step 3:Find half the coefficient of x and then square the result. Add this number to both sides of the equation. Step 4:Factor the resulting perfect square trinomial. Step 5:Use the square root property to solve the equation. Completing the Square

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 13 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve by completing the square. y 2 + 6y = –8 y 2 + 6y + 9 = –8 + 9 (y + 3) 2 = 1 y = – 3 ± 1 y = – 4 or – 2 y + 3 = ± =± 1 Solving Equations Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 14 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve by completing the square. y 2 + y – 7 = 0 y 2 + y = 7 y 2 + y + ¼ = 7 + ¼ (y + ½) 2 = Solving Equations Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 15 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve by completing the square. 2x 2 + 14x – 1 = 0 2x 2 + 14x = 1 x 2 + 7x = x 2 + 7x + = + (x + ) 2 = Solving Equations Example 1 2 1 2 7 2

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 16.3 Solving Quadratic Equations by the Quadratic Formula

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 17 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 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. We will not develop the formula now.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 18 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. If a, b, and c are real numbers and a ≠ 0, a quadratic equation written in the standard form ax 2 + bx + c = 0 has solutions The Quadratic Formula

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 19 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve 11n 2 – 9n = 1 by the quadratic formula. 11n 2 – 9n – 1 = 0, so a = 11, b = – 9, c = – 1 The Quadratic Formula Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 20 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. x 2 + 8x – 20 = 0 Multiply both sides by 8. a = 1, b = 8, c = –20 Solve x 2 + x – = 0 by the quadratic formula. The Quadratic Formula Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 21 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve x(x + 6) = –30 by the quadratic formula. x 2 + 6x + 30 = 0 a = 1, b = 6, c = 30 There is no real solution. The Quadratic Formula Example

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Quadratic Equations When Solving Quadratic Equations: 1)If the equation is in the form (ax + b) 2 = c, use the square root property to solve. 2)If not solved in step 1, write the equation in standard form. 3)Try to solve by factoring. 4)If you haven’t solved it yet, use the quadratic formula.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 23 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Approximate the exact solutions of 12x = 4x 2 + 4. Round to the nearest tenth. 0 = 4x 2 – 12x + 4 0 = 4(x 2 – 3x + 1) Let a = 1, b = –3, c = 1 Approximate Solutions to Quadratic Equations Example The approximate solutions are 0.4 and 2.6 to the nearest tenth. ≈ 0.4, 2.6

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 16.4 Graphing Quadratic Equations in Two Variables

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 25 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. We graphed linear equations in chapter 3. The graph of a quadratic equation is a parabola. The highest point or lowest point on the parabola is the vertex. The line of symmetry is the line that runs through the vertex and through the middle of the parabola. Graphs of Quadratic Equations

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 26 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. x y Graph: y = 2x 2 – 4 x y 0 –4–4 1 –2–2 –1–1 –2–2 24 –2–24 (2, 4) (–2, 4) (1, –2)(–1, – 2) (0, –4) Graphs of Quadratic Equations Example vertex lowest point

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 27 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Recall that: To find x-intercepts, let y = 0 and solve for x. To find y-intercepts, let x = 0 and solve for y. Helpful Hint

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 28 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. The vertex of the parabola y = ax 2 + bx + c has x-coordinate The corresponding y-coordinate of the vertex is obtained by substituting the x-coordinate into the equation and finding y. Vertex Formula – b 2a2a

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 29 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Graphs of Quadratic Equations Graphing Parabolas Defined by y = ax 2 + bx + c 1.Find the vertex by using the formula Don’t forget to find the y-value of the vertex. 2.Find the intercepts. Let x = 0 and solve for y to find the y-intercept. There will be only one. Let y = 0 and solve for x to find the x-intercept. There may be 0, 1, or 2. 3.Plot the vertex and the intercepts. 4.Find and plot additional points on the graph. Then draw a smooth curve through the plotted points. Keep in mind that if a > 0, the parabola opens upward and that if a < 0, the parabola opens downward.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 30 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. x y Graph: y = – 2x 2 + 4x + 5 x y 1 7 2 5 05 3–1 (3, –1) (–1, –1) (2, 5) (0, 5) (1, 7) Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex is: Graphs of Quadratic Equations Example vertex