Solve using the Square Root Method: 3(2x – 3)2 – 42 = 0?

2009-04-20 12:00 pm

回答 (8)

2009-04-20 12:06 pm
✔ 最佳答案
So (2x-3)^2 = 14

2x-3 = +/- sqrt 14

x = (3 +/- sqrt 14)/2
2009-04-20 12:08 pm
3(2x - 3)² - 42 = 0
3(4x² - 6x - 6x + 9) = 42
4x² - 12x + 9 = 14
4x² - 12x = 5
x² - 3x = 5/4
x² - 3/2x = 5/4 + (- 3/2)²
x² - 3/2x = 5/4 + 9/4
(x - 3/2)² = 7/2
x - 1.5 = 1.8708287

x = 1.8708287 + 1.5, x = 3.3708287
x = - 1.8708287 + 1.5, x = - 0.3708287

Answer: x = 3.3708287, - 0.3708287
2016-10-25 1:49 pm
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2009-04-20 12:22 pm
3(2x - 3)^2 - 42 = 0
3(2x - 3)^2 = 42
(2x - 3)^2 = 42/3
(2x - 3)^2 = 14
2x - 3 = ±√14
2x = 3 ±√14
x = (3 ±√14)/2
2009-04-20 12:14 pm
x = (3 - sqrt 14)/2
2009-04-20 12:11 pm
3(2x – 3)² – 42 = 0
3(2x – 3)² = 42
(2x – 3)² = 42/3 = 14
[√(2x – 3)²] = √14
(2x – 3) = ± √14
2x = 3 ± √14
x = (3 ± √14)/2
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2009-04-20 12:09 pm
(2x-3)^2=14
2x-3=3.74 , x=3.37
2x-3=-3.74, x=-0.18541
2009-04-20 12:05 pm
2x-3 = sqrt(14) --->x = (3+sqrt14)/2 and
2x-3 = -sqrt(14) --->x = (3 - sqrt14)/2


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