x^2 - 4x -12 = 0
(x + 2) (x - 6) = 0
x + 2 = 0 or x - 6 = 0
x + 2 - 2 = 0 -2 or x - 6 + 6 = 0 + 6
x = -2 or x = 6
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Limx→0 [ 1 / (x - 4) + 1 / (x + 4) ] / x = Limx→0 1 / (x2 - 4x) + 1 / (x2 + 4x) = Limx→0 (x2 + 4x) / (x4 - 16x2) + (x2 - 4x) / (x4 - 16x2) = Limx→0 (x2 + 4x - 4x + x2) / (x4 - 16x2) = Limx→0 2x2 / (x4 - 16x2) = Limx→0 2 / (x2 - 16) = 2 / (0 - 16) = -1/8
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
An equation with the solution set 1 and 3 can be written in factored form as (x-1)(x-3) = 0. When expanded, this equation becomes x^2 - 4x + 3 = 0. Therefore, the equation x^2 - 4x + 3 = 0 has the solution set 1 and 3.
Their sum is 4.
Yes. x2-4x+4=0 (x-2)(x-2)=0 x=2