Solving more quadratic equations part four



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Solving more quadratic equations part four



Solve 10x² - 19x + 6 = 0 for x.

First method

10x² - 19x + 6 = 0

(2x - 3)(5x - 2) = 0

x = 3 / 2; 2 / 5

 

Second method

10x² - 19x + 6 = 0

400x² - 760x + 240 = 0

400x² - 760x + 361 = 121

(20x - 19)² = 121

20x - 19 = ±11

20x = 19 ± 11

x = (19 ± 11) / 20

x = 3 / 2; 2 / 5

 

Third method

10x² - 19x + 6 = 0

a = 10, b = -19 and c = 6

b² - 4ac = 121

x = (19 ± √121) / 20

x = (19 ± 11) / 20

x = 3 / 2; 2 / 5

 

Solve 10x² - 21x + 9 = 0 for x.

First method

10x² - 21x + 9 = 0

(2x - 3)(5x - 3) = 0

x = 3 / 2; 3 / 5

 

Second method

10x² - 21x + 9 = 0

400x² - 840x + 360 = 0

400x² - 840x + 441 = 81

(20x - 21)² = 81

20x - 21 = ±9

20x = 21 ± 9

x = (21 ± 9) / 20

x = 3 / 2; 3 / 5

 

Third method

10x² - 21x + 9 = 0

a = 10, b = -21 and c = 9

b² - 4ac = 81

x = (21 ± √81) / 20

x = (21 ± 9) / 20

x = 3 / 2; 3 / 5

 

Solve 15x² - 31x + 10 = 0 for x.

First method

15x² - 31x + 10 = 0

(3x - 5)(5x - 2) = 0

x = 5 / 3; 2 / 5

 

Second method

15x² - 31x + 10 = 0

900x² - 1,860x + 600 = 0

900x² - 1,860x + 961 = 361

(30x - 31)² = 361

30x - 31 = ±19

30x = 31 ± 19

x = (31 ± 19) / 30

x = 5 / 3; 2 / 5

 

Third method

15x² - 31x + 10 = 0

a = 15, b = -31 and c = 10

b² - 4ac = 361

x = (31 ± √361) / 30

x = (31 ± 19) / 30

x = 5 / 3; 2 / 5


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0 user (users) favorited this work
  • View Count:289
  • Rating:General - Intended for all ages.
  • Publish Time:2021-12-01 09:39