Download The Quadratic Formula Joseph Lee Metropolitan Community College

The Quadratic Formula
Joseph Lee
Metropolitan Community College
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Proof.
ax 2 + bx + c = 0
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Proof.
ax 2 + bx + c = 0
ax 2 + bx
Joseph Lee
= −c
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Proof.
ax 2 + bx + c = 0
ax 2 + bx
= −c
ax 2 + bx
a
=
Joseph Lee
−c
a
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Proof.
ax 2 + bx + c = 0
ax 2 + bx
= −c
ax 2 + bx
a
=
b
x2 + x
a
Joseph Lee
−c
a
=−
c
a
The Quadratic Formula
Theorem: Quadratic Formula
For any equation ax 2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b 2 − 4ac
.
x=
2a
Proof.
ax 2 + bx + c = 0
ax 2 + bx
= −c
ax 2 + bx
a
=
−c
a
b
c
x2 + x
=−
a
a
2
2
b
b
c
b
2
x + x+
=− +
a
2a
a
2a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
b
b2
c
b2
x2 + x + 2 = − + 2
a
4a
a 4a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
b
b2
c
b2
x2 + x + 2 = − + 2
a
4a
a 4a
4ac
b 2
b2
=− 2 + 2
x+
2a
4a
4a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
b
b2
c
b2
x2 + x + 2 = − + 2
a
4a
a 4a
4ac
b 2
b2
=− 2 + 2
x+
2a
4a
4a
b
x+
2a
2
Joseph Lee
=
b2
4ac
− 2
2
4a
4a
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
b
b2
c
b2
x2 + x + 2 = − + 2
a
4a
a 4a
4ac
b 2
b2
=− 2 + 2
x+
2a
4a
4a
b
x+
2a
2
b
x+
2a
2
Joseph Lee
=
b2
4ac
− 2
2
4a
4a
=
b 2 − 4ac
4a2
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
s
r
b 2
b 2 − 4ac
x+
=±
2a
4a2
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
s
r
b 2
b 2 − 4ac
x+
=±
2a
4a2
√
b
b 2 − 4ac
=± √
x+
2a
4a2
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
s
r
b 2
b 2 − 4ac
x+
=±
2a
4a2
√
b
b 2 − 4ac
=± √
x+
2a
4a2
√
b
b 2 − 4ac
x+
=±
2a
2a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
s
r
b 2
b 2 − 4ac
x+
=±
2a
4a2
√
b
b 2 − 4ac
=± √
x+
2a
4a2
√
b
b 2 − 4ac
x+
=±
2a
2a
√
b 2 − 4ac
b
x =− ±
2a
2a
Joseph Lee
The Quadratic Formula
Theorem: Quadratic Formula (Continued)
s
r
b 2
b 2 − 4ac
x+
=±
2a
4a2
√
b
b 2 − 4ac
=± √
x+
2a
4a2
√
b
b 2 − 4ac
x+
=±
2a
2a
√
b 2 − 4ac
b
x =− ±
2a
2a
√
−b ± b 2 − 4ac
x =
2a
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
x =
−b ±
√
b 2 − 4ac
2a
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
√
b 2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x =
−b ±
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
√
b 2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x =
=
−b ±
−2 ±
√
4 + 32
2
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
√
b 2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x =
=
=
−b ±
−2 ±
√
4 + 32
2
√
−2 ± 36
2
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
√
b 2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x =
=
=
−b ±
−2 ±
√
4 + 32
2
√
−2 ± 36
−2 ± 6
=
2
2
Joseph Lee
The Quadratic Formula
Example 1
Solve.
x 2 + 2x − 8 = 0
Solution.
√
b 2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x =
=
−b ±
−2 ±
√
4 + 32
2
√
−2 ± 36
−2 ± 6
=
2
2
Thus, x = 2 or x = −4.
=
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
−b ±
√
b 2 − 4ac
2a
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
=
−b ±
5±
√
b 2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
=
=
−b ±
5±
5±
√
b 2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
√
25 + 24
6
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
=
=
=
−b ±
5±
5±
5±
√
b 2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
√
√
25 + 24
6
49
6
Joseph Lee
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
=
=
=
−b ±
5±
5±
5±
√
b 2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
√
√
25 + 24
6
49
6
Joseph Lee
=
5±7
6
The Quadratic Formula
Example 2
Solve.
3x 2 − 5x − 2 = 0
Solution.
x =
=
=
=
Thus, x = 2 or x =
−b ±
5±
5±
5±
√
b 2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
√
√
25 + 24
6
49
6
=
5±7
6
− 13 .
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Solution.
x =
−b ±
√
b 2 − 4ac
2a
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Solution.
x =
=
−b ±
3±
√
b 2 − 4ac
2a
p
(−3)2 − 4(1)(−7)
2(1)
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Solution.
x =
=
=
−b ±
3±
3±
√
b 2 − 4ac
2a
p
(−3)2 − 4(1)(−7)
2(1)
√
9 + 28
2
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Solution.
x =
=
−b ±
3±
3±
√
b 2 − 4ac
2a
p
(−3)2 − 4(1)(−7)
2(1)
√
9 + 28
2
√
3 ± 37
=
2
=
Joseph Lee
The Quadratic Formula
Example 3
Solve.
x 2 − 3x − 7 = 0
Solution.
x =
=
−b ±
3±
3±
√
b 2 − 4ac
2a
p
(−3)2 − 4(1)(−7)
2(1)
√
9 + 28
2
√
3 ± 37
=
2
√
√
3 + 37
3 − 37
Thus, x =
or x =
.
2
2
=
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
x =
−b ±
√
b 2 − 4ac
2a
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
√
b 2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x =
−b ±
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
√
b 2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x =
=
−b ±
−1 ±
√
1−8
4
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
√
b 2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x =
=
=
−b ±
−1 ±
√
1−8
4
√
−1 ± −7
4
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
√
b 2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x =
=
−b ±
−1 ±
√
1−8
4
√
√
−1 ± −7
−1 ± i 7
=
=
4
4
Joseph Lee
The Quadratic Formula
Example 4
Solve.
2x 2 + x + 1 = 0
Solution.
√
b 2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x =
=
−b ±
−1 ±
√
1−8
4
√
√
−1 ± −7
−1 ± i 7
=
=
4
4
√
√
−1 + i 7
−1 − i 7
Thus, x =
or x =
.
4
4
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
x =
7±
p
(−7)2 − 4(4)(3)
2(4)
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
x =
=
7±
7±
p
(−7)2 − 4(4)(3)
2(4)
√
49 − 48
8
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
x =
=
7±
7±
p
(−7)2 − 4(4)(3)
2(4)
√
49 − 48
8
√
7± 1
=
8
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
x =
=
7±
7±
p
(−7)2 − 4(4)(3)
2(4)
√
49 − 48
8
√
7± 1
7±1
=
=
8
8
Joseph Lee
The Quadratic Formula
Example 5
Solve.
4x 2 = 7x − 3
Solution.
4x 2 = 7x − 3
4x 2 − 7x + 3 = 0
x =
=
7±
7±
p
(−7)2 − 4(4)(3)
2(4)
√
49 − 48
8
√
7± 1
7±1
=
=
8
8
Thus, x = 1 or x = 43 .
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
x =
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
x =
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
√
−3 ± 9 − 4
=
−2
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
x =
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
√
−3 ± 9 − 4
=
−2
√
−3 ± 5
=
−2
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
x =
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
√
−3 ± 9 − 4
=
−2
√
√
−3 ± 5
3∓ 5
=
=
−2
2
Joseph Lee
The Quadratic Formula
Example 6
Solve.
3x − x 2 = 1
Solution.
3x − x 2 = 1
−x 2 + 3x − 1 = 0
x =
Thus, x =
√
3− 5
2
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
√
−3 ± 9 − 4
=
−2
√
√
−3 ± 5
3∓ 5
=
=
−2
2
or x =
√
3+ 5
2 .
Joseph Lee
The Quadratic Formula