# Chapter 9 Quadratic Equations Chapter Sections 9 1

• Slides: 33

Chapter Sections 9. 1 – Solving Quadratic Equations by the Square Root Property 9. 2 – Solving Quadratic Equations by Completing the Square 9. 3 – Solving Quadratic Equations by the Quadratic Formula 9. 4 – Graphing Quadratic Equations in Two Variables Martin-Gay, Introductory Algebra, 3 ed 2

§ 9. 1 Solving Quadratic Equations by the Square Root Property

Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a 2 = b, then Martin-Gay, Introductory Algebra, 3 ed 4

Square Root Property Example Solve x 2 = 49 Solve 2 x 2 = 4 x 2 = 2 Solve (y – 3)2 = 4 y=3 2 y = 1 or 5 Martin-Gay, Introductory Algebra, 3 ed 5

Square Root Property Example Solve x 2 + 4 = 0 x 2 = 4 There is no real solution because the square root of 4 is not a real number. Martin-Gay, Introductory Algebra, 3 ed 6

Square Root Property Example Solve (x + 2)2 = 25 x = 2 ± 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7 Martin-Gay, Introductory Algebra, 3 ed 7

Square Root Property Example Solve (3 x – 17)2 = 28 3 x – 17 = Martin-Gay, Introductory Algebra, 3 ed 8

§ 9. 2 Solving Quadratic Equations by Completing the Square

Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x 2 term is 1, as in our previous examples). Martin-Gay, Introductory Algebra, 3 ed 10

Completing the Square Example What constant term should be added to the following expressions to create a perfect square trinomial? x 2 – 10 x add 52 = 25 x 2 + 16 x add 82 = 64 x 2 – 7 x add Martin-Gay, Introductory Algebra, 3 ed 11

Completing the Square Example 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. Martin-Gay, Introductory Algebra, 3 ed 12

Completing the Square Solving a Quadratic Equation by Completing a Square 1) If the coefficient of x 2 is NOT 1, divide both sides of the equation by the coefficient. 2) Isolate all variable terms on one side of the equation. 3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation). 4) Factor the resulting trinomial. 5) Use the square root property. Martin-Gay, Introductory Algebra, 3 ed 13

Solving Equations Example Solve by completing the square. y 2 + 6 y = 8 y 2 + 6 y + 9 = 8 + 9 (y + 3)2 = 1 y+3=± =± 1 y = 3 ± 1 y = 4 or 2 Martin-Gay, Introductory Algebra, 3 ed 14

Solving Equations Example Solve by completing the square. y 2 + y – 7 = 0 y 2 + y = 7 y 2 + y + ¼ = 7 + ¼ (y + ½)2 = Martin-Gay, Introductory Algebra, 3 ed 15

Solving Equations Example Solve by completing the square. 2 x 2 + 14 x – 1 = 0 2 x 2 + 14 x = 1 x 2 + 7 x = ½ x 2 + 7 x + (x + =½+ = )2 = Martin-Gay, Introductory Algebra, 3 ed 16

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. Martin-Gay, Introductory Algebra, 3 ed 18

The Quadratic Formula A quadratic equation written in standard form, ax 2 + bx + c = 0, has the solutions. Martin-Gay, Introductory Algebra, 3 ed 19

The Quadratic Formula Example Solve 11 n 2 – 9 n = 1 by the quadratic formula. 11 n 2 – 9 n – 1 = 0, so a = 11, b = -9, c = -1 Martin-Gay, Introductory Algebra, 3 ed 20

The Quadratic Formula Example Solve x 2 + x – = 0 by the quadratic formula. x 2 + 8 x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20 Martin-Gay, Introductory Algebra, 3 ed 21

The Quadratic Formula Example Solve x(x + 6) = 30 by the quadratic formula. x 2 + 6 x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution. Martin-Gay, Introductory Algebra, 3 ed 22

The Discriminant The expression under the radical sign in the formula (b 2 – 4 ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively. Martin-Gay, Introductory Algebra, 3 ed 23

The Discriminant Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4 x + 12 x 2 = 0 a = 12, b = – 4, and c = 5 b 2 – 4 ac = (– 4)2 – 4(12)(5) = 16 – 240 = – 224 There are no real solutions. Martin-Gay, Introductory Algebra, 3 ed 24

Solving Quadratic Equations Steps in 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, Introductory Algebra, 3 ed 25

Solving Equations Example Solve 12 x = 4 x 2 + 4. 0 = 4 x 2 – 12 x + 4 0 = 4(x 2 – 3 x + 1) Let a = 1, b = -3, c = 1 Martin-Gay, Introductory Algebra, 3 ed 26

Solving Equations Example Solve the following quadratic equation. Martin-Gay, Introductory Algebra, 3 ed 27

§ 9. 4 Graphing Quadratic Equations in Two Variables

Graphs of Quadratic Equations We spent a lot of time graphing 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. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola. Martin-Gay, Introductory Algebra, 3 ed 29

Graphs of Quadratic Equations Example y Graph y = 2 x 2 – 4. x y 2 1 0 – 1 – 2 4 – 2 – 4 – 2 4 (– 2, 4) (2, 4) x (– 1, – 2) (0, – 4) Martin-Gay, Introductory Algebra, 3 ed 30

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. Martin-Gay, Introductory Algebra, 3 ed 31

Characteristics of the Parabola If the quadratic equation is written in standard form, y = ax 2 + 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. Martin-Gay, Introductory Algebra, 3 ed 32

Graphs of Quadratic Equations Example Graph y = – 2 x 2 + 4 x + 5. Since a = – 2 and b = 4, the graph opens down and the x-coordinate of the vertex is x y 3 – 1 2 5 1 7 0 5 – 1 y (0, 5) (– 1, – 1) Martin-Gay, Introductory Algebra, 3 ed (1, 7) (2, 5) (3, – 1) x 33