the quadratic formula and the discriminant
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The Quadratic Formula and the Discriminant. Algebra 2HN. Derive the quadratic formula from ax 2 + bx + c = 0 a ≠ 0. General form of a quadratic equation. Divide all by a. Simplify. Subtract c/a on both sides. Multiply by ½ and square the result. - PowerPoint PPT PresentationTRANSCRIPT
The Quadratic Formula and the Discriminant
Algebra 2HN
Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
a
cx
a
bx
a
c
a
bxx
aa
c
a
bx
a
ax
cbxax
22
2
2
2
2
1
0
0
0 General form of a quadratic equation.
Divide all by a
Simplify
Subtract c/a on both sides.
Multiply by ½ and square the result.
Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
2
22
2
22
2
22
222
4
4
2
4
4
42
42
22
a
acb
a
bx
a
a
a
c
a
b
a
bx
a
c
a
b
a
bx
a
b
a
c
a
bx
a
bx
Add the result to both sides.
Simplify
Multiply by common denominator
Simplify
Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
a
acbbx
a
acb
a
b
a
b
a
bx
a
acb
a
bx
a
acb
a
bx
2
4
2
4
222
2
4
2
4
4
2
2
2
2
2
22
Square root both sides
Simplify
Common denominator/subtract from both sides
Simplify
Quadratic Formula
The solutions of a quadratic equation of the form ax2 + bx + c with a ≠ 0 are given by this formula:
a
acbbx
2
42
MEMORIZE!!!!
Ex. 1: Solve t2 – 3t – 28 = 0
a = 1 b = -3 c = -28
42
8
2
113
72
14
2
1132
1132
1213
x
x
x
x
2
11293
)1(2
)28)(1(4)3()3(
2
4
2
2
x
x
a
acbbx
There are 2 distinct roots—Real and rational.
Ex. 1: Solve t2 – 3t – 28 = 0
CHECK:
t2 – 3t – 28 = 0
72 – 3(7) – 28 = 0
49 – 21 – 28 = 0
49 – 49 = 0
CHECK:
t2 – 3t – 28 = 0
(-4)2 – 3(-4) – 28 = 0
16 + 12 – 28 = 0
28 – 28 = 0
Ex. 1: Solve t2 – 3t – 28 = 0 -- GRAPH4
2
-2
-4
-6
-8
-10
-5 5 10 15
Ex. 2: Solve x2 – 8x + 16 = 0
a = 1 b = -8 c = 16
42
82
082
08
x
x
x
2
64648
)1(2
)16)(1(4)8()8(
2
4
2
2
x
x
a
acbbx
There is 1 distinct root—Real and rational.
Ex. 2: Solve x2 – 8x + 16 = 0
CHECK:
x2 – 8x + 16 = 0
(4)2 – 8(4) + 16 = 0
16 – 32 + 16 = 0
32 – 32 = 0
There is 1 distinct root—Real and rational.
Ex. 2: Solve Solve x2 – 8x + 16 = 0 -- GRAPH
8
6
4
2
-2
-4
-6
-8
5 10 15 20 25 30
Ex. 3: Solve 3p2 – 5p + 9 = 0
a = 3 b = -5 c = 9
6
835
6
835
6
835
6
835
ix
ix
ix
x
6
108255
)3(2
)9)(3(4)5()5(
2
4
2
2
x
x
a
acbbx
There is 2 imaginary roots.
Ex. 3: Solve 3p2 – 5p + 9 = 020
18
16
14
12
10
8
6
4
2
-2
5 10 15 20 25 30 35 40
NOTICE THAT THE PARABOLA DOES NOT TOUCH THE X-AXIS.
Note:
These three examples demonstrate a pattern that is useful in determining the nature of the root of a quadratic equation. In the quadratic formula, the expression under the radical sign, b2 – 4ac is called the discriminant. The discriminant tells the nature of the roots of a quadratic equation.
DISCRIMINANT The discriminant will tell you about the nature of the
roots of a quadratic equation.
acb 42
Equation Value of the discriminant
Roots Nature of roots
t2 – 3t – 28 = 0 b2 – 4ac =
(-3)2 – 4(1)(-28) = 121{7, - 4} 2 real
roots
x2 – 8x + 16 = 0 b2 – 4ac =
(-8)2 – 4(1)(16) = 0{0} 1 real root
3p2 – 5p + 9 = 0 b2 – 4ac =
(-5)2 – 4(3)(9) = -83
2 complex roots
6
835 ix
Ex. 4: Find the value of the discriminant of each equation and then describe the nature of its roots.
2x2 + x – 3 = 0
a = 2 b = 1 c = -3
b2 – 4ac = (1)2 – 4(2)(-3)
= 1 + 24
= 25
The value of the discriminant is positive and a perfect square, so 2x2 + x – 3 = 0 has two real roots and they are rational.
Ex. 5: Find the value of the discriminant of each equation and then describe the nature of its roots.
x2 + 8 = 0
a = 1 b = 0 c = 8
b2 – 4ac = (0)2 – 4(1)(8)
= 0 – 32
= – 32
The value of the discriminant is negative, so x2 + 8 = 0 has two imaginary/complex roots.