Solve fof x (解ｘ）
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Quadratic Equations
2006-10-16 3:16 pm
回答 (3)
2006-10-16 3:30 pm
✔ 最佳答案
4a^2 - 4ax + x^2 - 9 = 0(4a^2 - 4ax + x^2) - 9 = 0
(2a - x)^2 - 3^2 = 0
(2a - x + 3)(2a - x - 3) = 0
2a - x + 3 = 0 or 2a - x - 3 = 0
x = 2a + 3 or x = 2a - 3
2006-10-16 5:52 pm
4a² - 4ax + x² - 9=0
(2a)² - 2(2a)(x) + x² - 3² = 0
(2a - x)² - 3² = 0
[(2a - x) + 3][(2a - x) - 3] = 0
(2a - x + 3)(2a - x - 3) = 0
(2a - x + 3) = 0 or (2a - x - 3) = 0
2a + 3 = x or 2a - 3 = x
x = 2a + 3 or x = 2a - 3
(2a)² - 2(2a)(x) + x² - 3² = 0
(2a - x)² - 3² = 0
[(2a - x) + 3][(2a - x) - 3] = 0
(2a - x + 3)(2a - x - 3) = 0
(2a - x + 3) = 0 or (2a - x - 3) = 0
2a + 3 = x or 2a - 3 = x
x = 2a + 3 or x = 2a - 3
2006-10-16 4:47 pm
4a²-4ax+x²-9=0
(2a-x)²-3²=0
[(2a-x)+3][(2a-x)-3]=0
( -x+2a+3)(-x+2a-3)=0
(x-2a-3)(x-2a+3)=0
[x-(2a+3)][x-(2a-3)]=0
x=2a+3 or x=2a-3
(2a-x)²-3²=0
[(2a-x)+3][(2a-x)-3]=0
( -x+2a+3)(-x+2a-3)=0
(x-2a-3)(x-2a+3)=0
[x-(2a+3)][x-(2a-3)]=0
x=2a+3 or x=2a-3
收錄日期: 2021-04-12 22:40:54
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