Download Quadratic Equations - Mayfield City Schools

Chapter 16
Quadratic
Equations
§ 16.2
Solving Quadratic
Equations by Completing
the Square
Completing the Square
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.
Martin-Gay, Developmental Mathematics
3
Solving Equations
Example
Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y+3=± 1=±1
y = 3 ± 1
y = 4 or 2
Martin-Gay, Developmental Mathematics
4
Solving Equations
Example
Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
(y +
½)2
=
29
4
1
29
29
y 

2
4
2
1
29  1  29
y 

2
2
2
Martin-Gay, Developmental Mathematics
5
Solving Equations
Example
Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
x2 + 7x +
49
4
=½+
49
4
=
51
4
7 2
51
(x + ) =
2
4
x
7
51
51


2
4
2
7
51  7  51
x 

2
2
2
Martin-Gay, Developmental Mathematics
6
§ 16.3
Solving Quadratic
Equations by the
Quadratic Formula
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
8
The Quadratic Formula
A quadratic equation written in standard
form, ax2 + bx + c = 0, has the solutions.
 b  b  4ac
x
2a
2
Martin-Gay, Developmental Mathematics
9
The Quadratic Formula
Example
Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
9  (9)  4(11)( 1) 9  81  44 9  125
n



22
22
2(11)
2
95 5
22
Martin-Gay, Developmental Mathematics
10
The Quadratic Formula
Example
1 2
5
Solve x + x – = 0 by the quadratic formula.
8
2
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
 8  (8) 2  4(1)( 20)  8  64  80  8  144
x



2(1)
2
2
 8  12 20
4

or ,  10 or 2
2
2
2
Martin-Gay, Developmental Mathematics
11
The Quadratic Formula
Example
Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
 6  (6)  4(1)(30)  6  36  120  6   84
x


2
2
2(1)
2
So there is no real solution.
Martin-Gay, Developmental Mathematics
12
Graphs of Quadratic Equations
Example
y
Graph y = 2x2 – 4.
x
y
2
4
1
–2
0
–4
–1
–2
–2
4
(–2, 4)
(2, 4)
x
(–1, – 2)
(1, –2)
(0, –4)
Martin-Gay, Developmental Mathematics
13
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, Developmental Mathematics
14
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.
b
2) the x-coordinate of the vertex is  .
2a
To find the corresponding y-coordinate, you
substitute the x-coordinate into the equation
and evaluate for y.
Martin-Gay, Developmental Mathematics
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Graphs of Quadratic Equations
Example
Graph y = –2x2 + 4x + 5.
Since a = –2 and b = 4, the
graph opens down and the
x-coordinate of the vertex
4
is 
1
y
(0, 5)
(1, 7)
(2, 5)
2(2)
x
y
3
–1
2
5
1
7
0
5
–1
–1
(–1, –1)
Martin-Gay, Developmental Mathematics
(3, –1)
x
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