solve 3x^2+4x-7=0 using the nul factor law???thanks?
回答 (5)
2008-03-14 10:39 am
✔ 最佳答案
if you have such an equation you see immediately adding the coefficients between power of x and the constant that their sum is 0 (3+4-7). So you have one root which is x=1.After this you remmeber product of roots =c/a here -7/3 and so the second root is x=-7/3
2008-03-14 6:20 pm
3x^2+4x-7=0
9x+4x=0+7
13x=7
therefore,x= -7/13
9x+4x=0+7
13x=7
therefore,x= -7/13
2008-03-14 6:18 pm
3x^2 + 4x - 7 = 0
(3x + 7)(x - 1) = 0
3x + 7 = 0
3x = -7
x = -7/3
x = -2 1/3
x - 1 = 0
x = 1
â´ x = -2 1/3 , 1
(3x + 7)(x - 1) = 0
3x + 7 = 0
3x = -7
x = -7/3
x = -2 1/3
x - 1 = 0
x = 1
â´ x = -2 1/3 , 1
2008-03-14 6:13 pm
the form of this question is that
a+b+c=0(a=3, b=4, c=-7)
so, x1=1 and x2=c/a=-7/3
a+b+c=0(a=3, b=4, c=-7)
so, x1=1 and x2=c/a=-7/3
2008-03-14 5:31 pm
3x^2+4x-7 = (3x+7)(x-1) = 0
so x=1, -7/3
ANSWER
so x=1, -7/3
ANSWER
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