Solving more quadratic equations part one



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Solving more quadratic equations part one



Solve 4x² - 16x + 15 = 0 for x.

First method

4x² - 16x + 15 = 0

(2x - 3)(2x - 5) = 0

x = 3 / 2; 5 / 2

 

Second method

4x² - 16x + 15 = 0

4x² - 16x + 16 = 1

(2x - 4)² = 1

2x - 4 = ±1

2x = 4 ± 1

x = (4 ± 1) / 2

x = 5 / 2; 3 / 2

 

Third method

4x² - 16x + 15 = 0

a = 4, b = -16 and c = 15

b² - 4ac = 16

x = (16 ± √16) / 8

x = (16 ± 4) / 8

x = (4 ± 1) / 2

x = 5 / 2; 3 / 2

 

Solve 6x² - 19x + 15 = 0 for x.

First method

6x² - 19x + 15 = 0

(2x - 3)(3x - 5) = 0

x = 3 / 2; 5 / 3

 

Second method

6x² - 19x + 15 = 0

144x² - 456x + 360 = 0

144x² - 456x + 361 = 1

(12x - 19)² = 1

12x - 19 = ±1

12x = 19 ± 1

x = (19 ± 1) / 12

x = 5 / 3; 3 / 2

 

Third method

6x² - 19x + 15 = 0

a = 6, b = -19 and c = 15

b² - 4ac = 1

x = (19 ± √1) / 12

x = (19 ± 1) / 12

x = 5 / 3; 3 / 2

 

Solve 6x² - 25x + 25 = 0 for x.

First method

6x² - 25x + 25 = 0

(2x - 5)(3x - 5) = 0

x = 5 / 2; 5 / 3

 

Second method

6x² - 25x + 25 = 0

144x² - 600x + 600 = 0

144x² - 600x + 625 = 25

(12x - 25)² = 25

12x - 25 = ±5

12x = 25 ± 5

x = (25 ± 5) / 12

x = 5 / 2; 5 / 3

 

Third method

6x² - 25x + 25 = 0

a = 6, b = -25 and c = 25

b² - 4ac = 25

x = (25 ± √25) / 12

x = (25 ± 5) / 12

x = 5 / 2; 5 / 3


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0 user (users) favorited this work
  • View Count:284
  • Rating:General - Intended for all ages.
  • Publish Time:2021-12-01 09:11