Does the quadratic formula always work for a quadratic? If it does, why are the factors sometimes imaginary numbers?
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- Yes, as long as your working on a field with characteristic $;\neq2;$ . (2) If you're working on the reals, that's because there don't exist square roots of negative real numbers within the real numbers field, but there are such within the complex numbers field.
– Timbuc Jan 10 '15 at 11:16
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To answer your question, yes, the formula always works for quadratic equations, because from the equation $ax^2+bx+c=0$, one can derive the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ manually. Here is a video on Youtube showing such a derivation.
The factors become imaginary numbers if $b^2-4ac<0$, which means $b^2-4ac$ is negative. And taking the square root of a negative number produces an imaginary number.
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Another reason why the factors are sometimes imaginary or complex is because not every parabola crosses the x-axis in the real plane. For an example, the equation $y=x^2 +1$ has as a vertex (and minimum) $(0,1)$ ,and goes up from there. So there cannot be any real roots. The actual roots ($\pm i$) are in the complex plane, and the fundamental theorem of algebra states that a polynomial always has a full complement of complex roots (Up to repetition)
Alan
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