Chapter 16

Quadratic Equations

§ 16.2

Solving Quadratic Equations by Completing

the Square

Martin-Gay, Developmental Mathematics 3

Solving a Quadratic Equation by Completing a Square

1) If the coefficient of x2 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.

Completing the Square

Martin-Gay, Developmental Mathematics 4

Solve by completing the square.

y2 + 6y = 8y2 + 6y + 9 = 8 + 9

(y + 3)2 = 1

y = 3 ± 1

y = 4 or 2

y + 3 = ± = ± 11

Solving Equations

Example

Martin-Gay, Developmental Mathematics 5

Solve by completing the square.

y2 + y – 7 = 0

y2 + y = 7

y2 + y + ¼ = 7 + ¼

2

29

4

29

2

1y

2

291

2

29

2

1 y

(y + ½)2 = 429

Solving Equations

Example

Martin-Gay, Developmental Mathematics 6

Solve by completing the square.

2x2 + 14x – 1 = 0

2x2 + 14x = 1

x2 + 7x = ½

2

51

4

51

2

7x

2

517

2

51

2

7 x

x2 + 7x + = ½ + = 4

49

4

49

4

51

(x + )2 = 4

51

2

7

Solving Equations

Example

§ 16.3

Solving Quadratic Equations by the

Quadratic Formula

Martin-Gay, Developmental Mathematics 8

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, Developmental Mathematics 9

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

a

acbbx

2

42

The Quadratic Formula

Martin-Gay, Developmental Mathematics 10

Solve 11n2 – 9n = 1 by the quadratic formula.

11n2 – 9n – 1 = 0, so

a = 11, b = -9, c = -1

)11(2

)1)(11(4)9(9 2

n

22

44819

22

1259

22

559

The Quadratic Formula

Example

Martin-Gay, Developmental Mathematics 11

)1(2

)20)(1(4)8(8 2

x

2

80648

2

1448

2

128 20 4 or , 10 or 22 2

x2 + 8x – 20 = 0 (multiply both sides by 8)

a = 1, b = 8, c = 20

8

1

2

5Solve x2 + x – = 0 by the quadratic formula.

The Quadratic Formula

Example

Martin-Gay, Developmental Mathematics 12

Solve x(x + 6) = 30 by the quadratic formula.

x2 + 6x + 30 = 0

a = 1, b = 6, c = 30

)1(2

)30)(1(4)6(6 2

x

2

120366

2

846

So there is no real solution.

The Quadratic Formula

Example

Martin-Gay, Developmental Mathematics 13

x

y

Graph y = 2x2 – 4.

x y

0 –4

1 –2

–1 –2

2 4

–2 4

(2, 4)(–2, 4)

(1, –2)(–1, – 2)

(0, –4)

Graphs of Quadratic Equations

Example

Martin-Gay, Developmental Mathematics 14

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.

Intercepts of the Parabola

Martin-Gay, Developmental Mathematics 15

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 . a

b

2

To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.

Characteristics of the Parabola

Martin-Gay, Developmental Mathematics 16

x

yGraph y = –2x2 + 4x + 5.

x y

1 7

2 5

0 5

3 –1

–1 –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 1

)2(2

4

Graphs of Quadratic Equations

Example