1/4 x^2 + x + 1 = 0 solve by factoring?
回答 (4)
2008-10-11 4:05 pm
✔ 最佳答案
1/4 x^2 + x + 1 = 0x^2 + 4x + 4 = 0
[x + 2]^2 = 0
x + 2 = sqrt {0} = 0
x = - 2
2008-10-11 10:49 pm
if you don't like the fraction in the beginning, you could always multiply the WHOLE equation by 4. That would give you:
x^2 + 4x + 4 = 0
That makes it a little easier to factor.
good luck
x^2 + 4x + 4 = 0
That makes it a little easier to factor.
good luck
2008-10-11 10:46 pm
1/4 x^2 + x + 1 = 0
(x+2)^2 = 0
x = -2
There is only one real root since the discriminant is zero.
(x+2)^2 = 0
x = -2
There is only one real root since the discriminant is zero.
2008-10-11 11:11 pm
1/4(x^2) + x + 1 = 0
x^2/4 + x + 1 = 0
4(x^2/4 + x + 1) = 4(0)
x^2 + 4x + 4 = 0
x^2 + 2x + 2x + 4 = 0
(x^2 + 2x) + (2x + 4) = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
x + 2 = 0
x = -2
â´ x = -2
x^2/4 + x + 1 = 0
4(x^2/4 + x + 1) = 4(0)
x^2 + 4x + 4 = 0
x^2 + 2x + 2x + 4 = 0
(x^2 + 2x) + (2x + 4) = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
x + 2 = 0
x = -2
â´ x = -2
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