chapter 16 quadratic equations. § 16.2 solving quadratic equations by completing the square
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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
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