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(4 ± i2) where i2 = -1
x + y = -2 ∴ y = -2 - x xy = 2 ∴ x(-2 - x) = 2 ∴ -2x - x2 = 2 ∴ x2 + 2x + 2 = 0 ∴ x2 + 2x + 1 = -1 ∴ (x + 1)2 = -1 ∴ x + 1 = i ∴ x = i - 1 recall: y = - 2 - x ∴ y = - 2 - (i - 1) ∴ y = -2 - i + 1 ∴ y = - i - 1 ∴ x3 + y3 = (i - 1)3 + (- i - 1)3 = (i - 1)(i2 - 2i + 1) + (- i - 1)(i2 + 2i + 2) = i3 - 2i2 + i - i2 + 2i - 1 - i3 - 2i2 - 2i - i2 - 2i - 2 = i3 - i3 - 2i2 - i2 - 2i2 - i2 + i + 2i - 2i - 2i - 1 - 2 = -6i2 - i - 3 = 3 - i
HI will be consumed. The reaction will proceed to the left. More I2 will form.
Cl2(g) + 2KI --> 2KCl(aq) + I2(s)
In math i=√(-1), in physics and engineering j=√(-1), and i2 or j2 =-1.