16x2-8x=1=0?
回答 (5)
2008-02-11 4:26 pm
✔ 最佳答案
[05]16x^2-8x+1=0
(4x)^2-2*4x*1+(1)^2=0
(4x-1)^2=0
4x-1=0
4x=1
x= 1/4
2008-02-11 4:23 pm
Assuming you meant:
16x^2 - 8x + 1 = 0
factor:
(4x - 1)(4x - 1) = 0
Since both terms are the same we can just set one term equal to 0:
4x - 1 = 0
4x = 1
x = 1/4.
16x^2 - 8x + 1 = 0
factor:
(4x - 1)(4x - 1) = 0
Since both terms are the same we can just set one term equal to 0:
4x - 1 = 0
4x = 1
x = 1/4.
2008-02-11 4:38 pm
if it is 16x² - 8x + 1 = 0 then
16x² - 4x -4x + 1 = 0
4x(4x - 1) -1 ( 4x - 1) = 0
(4x - 1) ( 4x - 1) = 0
(4x - 1)² = 0
4x - 1 = 0
4x = 1
x = 1/4.
16x² - 4x -4x + 1 = 0
4x(4x - 1) -1 ( 4x - 1) = 0
(4x - 1) ( 4x - 1) = 0
(4x - 1)² = 0
4x - 1 = 0
4x = 1
x = 1/4.
2008-02-11 4:32 pm
Do you mean: 16x^2 - 8x + 1 = 0
16x^2 - 8x + 1 = 0
(4x - 1)(4x - 1) = 0
4x - 1 = 0
4x = 1
x = 1/4
∴ x = 1/4
16x^2 - 8x + 1 = 0
(4x - 1)(4x - 1) = 0
4x - 1 = 0
4x = 1
x = 1/4
∴ x = 1/4
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