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Amy Ferguson Moncure

H Algebra 2/Trig · Unit 2

Quadratic Formula

Solve ax² + bx + c = 0 with the quadratic formula. Leave answers in simplest radical form, including i.

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Key Ideas Easy Medium A Medium B Hard Quiz

x = −b ± square root of b² − 4ac2a

Quadratic Formula
x = −b ± square root of b² − 4ac2a

x² + 2x − 3 = 0

a = 1
b = 2
c = −3

x = −2 ± square root of 2² − 4(1)(−3)2

x = −2 ± square root of 4 − (−12)2

x = −2 ± square root of 4 + 122

x = −2 ± square root of 162

x = −2 ± 42

x = −2 + 42

x = 22

x = 1

x = −2 − 42

x = −62

x = −3

x = 1 and x = −3

Careful!!
Careful when b is negative!!2x² − 3x − 1 = 0
a = 2
b = −3
c = −1

x = −b ± square root of b² − 4ac2a

x = −(−3) ± square root of (−3)² − 4(2)(−1)4

x = 3 ± square root of 9 − (−8)4

x = 3 ± square root of 9 + 84

x = 3 ± square root of 174

x = 3 ± square root of 174

Ready???
Is the Equation Ready???Set equation equal to zero first.x² = −4x + 1

x² = −4x + 1

x² + 4x − 1 = 0

a = 1
b = 4
c = −1

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(1)(−1)2

x = −4 ± square root of 16 − (−4)2

x = −4 ± square root of 16 + 42

x = −4 ± square root of 202

x = −4 ± 2square root of 52

x = −2 ± square root of 5

x = −2 ± square root of 5

Song
Quadratic Formula · Pop Goes the Weasel
Level 1

Easy

The equation is set equal to zero. Integer or fraction solutions.

1
Solve by the quadratic formula. x² + 4x + 3 = 0
Show Answer
a = 1
b = 4
c = 3

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(1)(3)2

x = −4 ± square root of 16 − 122

x = −4 ± square root of 42

x = −4 ± 22

x = −4 + 22

x = −22

x = −1

x = −4 − 22

x = −62

x = −3

Answer: x = −1 and x = −3

2
Solve by the quadratic formula. 2x² − 3x − 2 = 0
Show Answer
a = 2
b = −3
c = −2

x = −b ± square root of b² − 4ac2a

x = −(−3) ± square root of (−3)² − 4(2)(−2)4

x = 3 ± square root of 9 − (−16)4

x = 3 ± square root of 9 + 164

x = 3 ± square root of 254

x = 3 ± 54

x = 3 + 54

x = 84

x = 2

x = 3 − 54

x = −24

x = −12

Answer: x = 2 and x = −12

3
Solve by the quadratic formula. x² − 5x − 6 = 0
Show Answer
a = 1
b = −5
c = −6

x = −b ± square root of b² − 4ac2a

x = −(−5) ± square root of (−5)² − 4(1)(−6)2

x = 5 ± square root of 25 − (−24)2

x = 5 ± square root of 25 + 242

x = 5 ± square root of 492

x = 5 ± 72

x = 5 + 72

x = 122

x = 6

x = 5 − 72

x = −22

x = −1

Answer: x = 6 and x = −1

4
Solve by the quadratic formula. 3x² + 4x − 4 = 0
Show Answer
a = 3
b = 4
c = −4

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(3)(−4)6

x = −4 ± square root of 16 − (−48)6

x = −4 ± square root of 16 + 486

x = −4 ± square root of 646

x = −4 ± 86

x = −4 + 86

x = 46

x = 23

x = −4 − 86

x = −126

x = −2

Answer: x = 23 and x = −2

5
Solve by the quadratic formula. x² − 10x + 25 = 0
Show Answer
a = 1
b = −10
c = 25

x = −b ± square root of b² − 4ac2a

x = −(−10) ± square root of (−10)² − 4(1)(25)2

x = 10 ± square root of 100 − 1002

x = 10 ± square root of 02

x = 10 ± 02

x = 102

x = 5

Answer: x = 5

Level 2A

Medium A

The equation is set equal to zero. Two solutions with radicals.

1
Solve by the quadratic formula. x² + 4x − 1 = 0
Show Answer
a = 1
b = 4
c = −1

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(1)(−1)2

x = −4 ± square root of 16 − (−4)2

x = −4 ± square root of 16 + 42

x = −4 ± square root of 202

x = −4 ± 2square root of 52

x = −2 ± square root of 5

Answer: x = −2 ± square root of 5

2
Solve by the quadratic formula. x² − 6x + 4 = 0
Show Answer
a = 1
b = −6
c = 4

x = −b ± square root of b² − 4ac2a

x = −(−6) ± square root of (−6)² − 4(1)(4)2

x = 6 ± square root of 36 − 162

x = 6 ± square root of 202

x = 6 ± 2square root of 52

x = 3 ± square root of 5

Answer: x = 3 ± square root of 5

3
Solve by the quadratic formula. 2x² + 2x − 1 = 0
Show Answer
a = 2
b = 2
c = −1

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(2)(−1)4

x = −2 ± square root of 4 − (−8)4

x = −2 ± square root of 4 + 84

x = −2 ± square root of 124

x = −2 ± 2square root of 34

x = −1 ± square root of 32

Answer: x = −1 ± square root of 32

4
Solve by the quadratic formula. 3x² − 4x − 1 = 0
Show Answer
a = 3
b = −4
c = −1

x = −b ± square root of b² − 4ac2a

x = −(−4) ± square root of (−4)² − 4(3)(−1)6

x = 4 ± square root of 16 − (−12)6

x = 4 ± square root of 16 + 126

x = 4 ± square root of 286

x = 4 ± 2square root of 76

x = 2 ± square root of 73

Answer: x = 2 ± square root of 73

5
Solve by the quadratic formula. x² + 2x − 5 = 0
Show Answer
a = 1
b = 2
c = −5

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(1)(−5)2

x = −2 ± square root of 4 − (−20)2

x = −2 ± square root of 4 + 202

x = −2 ± square root of 242

x = −2 ± 2square root of 62

x = −1 ± square root of 6

Answer: x = −1 ± square root of 6

Level 2B

Medium B

The equation is set equal to zero. Imaginary solutions.

1
Solve by the quadratic formula. x² + 2x + 5 = 0
Show Answer
a = 1
b = 2
c = 5

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(1)(5)2

x = −2 ± square root of 4 − 202

x = −2 ± square root of −162

x = −2 ± isquare root of 162

x = −2 ± 4i2

x = −1 ± 2i

Answer: x = −1 ± 2i

2
Solve by the quadratic formula. x² − 6x + 13 = 0
Show Answer
a = 1
b = −6
c = 13

x = −b ± square root of b² − 4ac2a

x = −(−6) ± square root of (−6)² − 4(1)(13)2

x = 6 ± square root of 36 − 522

x = 6 ± square root of −162

x = 6 ± isquare root of 162

x = 6 ± 4i2

x = 3 ± 2i

Answer: x = 3 ± 2i

3
Solve by the quadratic formula. 2x² + 4x + 5 = 0
Show Answer
a = 2
b = 4
c = 5

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(2)(5)4

x = −4 ± square root of 16 − 404

x = −4 ± square root of −244

x = −4 ± isquare root of 244

x = −4 ± 2isquare root of 64

x = −2 ± isquare root of 62

Answer: x = −2 ± isquare root of 62

4
Solve by the quadratic formula. 3x² + 2x + 2 = 0
Show Answer
a = 3
b = 2
c = 2

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(3)(2)6

x = −2 ± square root of 4 − 246

x = −2 ± square root of −206

x = −2 ± isquare root of 206

x = −2 ± 2isquare root of 56

x = −1 ± isquare root of 53

Answer: x = −1 ± isquare root of 53

5
Solve by the quadratic formula. 2x² − 2x + 5 = 0
Show Answer
a = 2
b = −2
c = 5

x = −b ± square root of b² − 4ac2a

x = −(−2) ± square root of (−2)² − 4(2)(5)4

x = 2 ± square root of 4 − 404

x = 2 ± square root of −364

x = 2 ± isquare root of 364

x = 2 ± 6i4

x = 1 ± 3i2

Answer: x = 1 ± 3i2

Level 3

Hard

The equation is not set equal to zero. One, two, or zero real solutions.

1
Solve by the quadratic formula. 2x² + 5x = −2
Show Answer

2x² + 5x = −2

2x² + 5x + 2 = 0

a = 2
b = 5
c = 2

x = −b ± square root of b² − 4ac2a

x = −5 ± square root of 5² − 4(2)(2)4

x = −5 ± square root of 25 − 164

x = −5 ± square root of 94

x = −5 ± 34

x = −5 + 34

x = −24

x = −12

x = −5 − 34

x = −84

x = −2

Answer: x = −12 and x = −2

2
Solve by the quadratic formula. 2x² = 6x − 1
Show Answer

2x² = 6x − 1

2x² − 6x + 1 = 0

a = 2
b = −6
c = 1

x = −b ± square root of b² − 4ac2a

x = −(−6) ± square root of (−6)² − 4(2)(1)4

x = 6 ± square root of 36 − 84

x = 6 ± square root of 284

x = 6 ± 2square root of 74

x = 3 ± square root of 72

Answer: x = 3 ± square root of 72

3
Solve by the quadratic formula. x² − 4x + 7 = 3
Show Answer

x² − 4x + 7 = 3

x² − 4x + 4 = 0

a = 1
b = −4
c = 4

x = −b ± square root of b² − 4ac2a

x = −(−4) ± square root of (−4)² − 4(1)(4)2

x = 4 ± square root of 16 − 162

x = 4 ± square root of 02

x = 4 ± 02

x = 42

x = 2

Answer: x = 2

4
Solve by the quadratic formula. x² + 4x + 9 = 2x
Show Answer

x² + 4x + 9 = 2x

x² + 2x + 9 = 0

a = 1
b = 2
c = 9

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(1)(9)2

x = −2 ± square root of 4 − 362

x = −2 ± square root of −322

x = −2 ± isquare root of 322

x = −2 ± 4isquare root of 22

x = −1 ± 2isquare root of 2

Answer: x = −1 ± 2isquare root of 2

5
Solve by the quadratic formula. 2x² + 3x = 2
Show Answer

2x² + 3x = 2

2x² + 3x − 2 = 0

a = 2
b = 3
c = −2

x = −b ± square root of b² − 4ac2a

x = −3 ± square root of 3² − 4(2)(−2)4

x = −3 ± square root of 9 − (−16)4

x = −3 ± square root of 9 + 164

x = −3 ± square root of 254

x = −3 ± 54

x = −3 + 54

x = 24

x = 12

x = −3 − 54

x = −84

x = −2

Answer: x = 12 and x = −2

Quiz

Quiz

A mix from every level. Use this to see if you are ready.

1
Solve by the quadratic formula. 2x² − 7x + 3 = 0
Show Answer
a = 2
b = −7
c = 3

x = −b ± square root of b² − 4ac2a

x = −(−7) ± square root of (−7)² − 4(2)(3)4

x = 7 ± square root of 49 − 244

x = 7 ± square root of 254

x = 7 ± 54

x = 7 + 54

x = 124

x = 3

x = 7 − 54

x = 24

x = 12

Answer: x = 3 and x = 12

2
Solve by the quadratic formula. x² + 2x − 4 = 0
Show Answer
a = 1
b = 2
c = −4

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(1)(−4)2

x = −2 ± square root of 4 − (−16)2

x = −2 ± square root of 4 + 162

x = −2 ± square root of 202

x = −2 ± 2square root of 52

x = −1 ± square root of 5

Answer: x = −1 ± square root of 5

3
Solve by the quadratic formula. x² − 10x + 28 = 3
Show Answer

x² − 10x + 28 = 3

x² − 10x + 25 = 0

a = 1
b = −10
c = 25

x = −b ± square root of b² − 4ac2a

x = −(−10) ± square root of (−10)² − 4(1)(25)2

x = 10 ± square root of 100 − 1002

x = 10 ± square root of 02

x = 10 ± 02

x = 102

x = 5

Answer: x = 5

4
Solve by the quadratic formula. x² − 4x + 13 = 0
Show Answer
a = 1
b = −4
c = 13

x = −b ± square root of b² − 4ac2a

x = −(−4) ± square root of (−4)² − 4(1)(13)2

x = 4 ± square root of 16 − 522

x = 4 ± square root of −362

x = 4 ± isquare root of 362

x = 4 ± 6i2

x = 2 ± 3i

Answer: x = 2 ± 3i

5
Solve by the quadratic formula. 3x² = 4x + 4
Show Answer

3x² = 4x + 4

3x² − 4x − 4 = 0

a = 3
b = −4
c = −4

x = −b ± square root of b² − 4ac2a

x = −(−4) ± square root of (−4)² − 4(3)(−4)6

x = 4 ± square root of 16 − (−48)6

x = 4 ± square root of 16 + 486

x = 4 ± square root of 646

x = 4 ± 86

x = 4 + 86

x = 126

x = 2

x = 4 − 86

x = −46

x = −23

Answer: x = 2 and x = −23

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