For a, b, c ∈ R and a ≠ 0, let ax2 + bx + c = 0 be given. If Δ=b2-4ac < 0, there is no solution set on R. For example, lets investigate the equation x2+9=0. To find the solution set, we get xkare=-9, then x1=-√-9 and x2=√-9. Then the solution is empty on R since √-9 ∉ R.
In other words, in the equation x2+9=0, for a=1, b=0, c=9, Δ=b2-4ac = 02 - 4.1.9 = -36 < 0, therefore we need a new number set which contains the set of real numbers. This set is called set of complex numbers. It is shown as C. Also √-9 ∈ C.
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The imaginary number "i" is defined as the number whose square is -1. (√-1=i). So, √-9 = 3.√-1 = 3i
The roots of the equation x2+9 = 0 are x1=-√-9 so x1=-3i and x2=√-9 which is x2=3i. SS={-3i, 3i}.
