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Solving more quadratic equations part three



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  • Publish Time:2021-12-01 09:29

Solving more quadratic equations part three



Solve 6x² - 19x + 10 = 0 for x.

First method

6x² - 19x + 10 = 0

(2x - 5)(3x - 2) = 0

x = 5 / 2; 2 / 3

 

Second method

6x² - 19x + 10 = 0

144x² - 456x + 240 = 0

144x² - 456x + 361 = 121

(12x - 19)² = 121

12x - 19 = ±11

12x = 19 ± 11

x = (19 ± 11) / 12

x = 5 / 2; 2 / 3

 

Third method

6x² - 19x + 10 = 0

a = 6, b = -19 and c = 10

b² - 4ac = 121

x = (19 ± √121) / 12

x = (19 ± 11) / 12

x = 5 / 2; 2 / 3

 

Solve 9x² - 21x + 10 = 0 for x.

First method

9x² - 21x + 10 = 0

(3x - 2)(3x - 5) = 0

x = 2 / 3; 5 / 3

 

Second method

9x² - 21x + 10 = 0

36x² - 84x + 40 = 0

36x² - 84x + 49 = 9

(6x - 7)² = 9

6x - 7 = ±3

6x = 7 ± 3

x = (7 ± 3) / 6

x = 5 / 3; 2 / 3

 

Third method

9x² - 21x + 10 = 0

a = 9, b = -21 and c = 10

b² - 4ac = 81

x = (21 ± √81) / 18

x = (21 ± 9) / 18

x = (7 ± 3) / 6

x = 5 / 3; 2 / 3

 

Solve 10x² - 31x + 15 = 0 for x.

First method

10x² - 31x + 15 = 0

(2x - 5)(5x - 3) = 0

x = 5 / 2; 3 / 5

 

Second method

10x² - 31x + 15 = 0

400x² - 1,240x + 600 = 0

400x² - 1,240x + 961 = 361

(20x - 31)² = 361

20x - 31 = ±19

20x = 31 ± 19

x = (31 ± 19) / 20

x = 5 / 2; 3 / 5

 

Third method

10x² - 31x + 15 = 0

a = 10, b = -31 and c = 15

b² - 4ac = 361

x = (31 ± √361) / 20

x = (31 ± 19) / 20

x = 5 / 2; 3 / 5


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0 user (users) favorited this work
  • View Count:494
  • Rating:General - Intended for all ages.
  • Publish Time:2021-12-01 09:29