Question 139569
I presume by 'D' you mean the determinant in the quadratic equation.  If D is less than zero, then you are faced with the problem of taking the square root of a negative number.  If you think about that for a moment, you will realize it is an impossible situation.  Taking a square root means finding two equal factors for a number and if the factors are equal, they must have the same sign.  If two numbers with the same sign are multiplied together, you always get a positive result.  So now what?


Mathematicians invented a special number, called an imaginary number, <i>i</i>, that is defined thusly: {{{i^2=-1}}}.  Turns out that this is the only imaginary number needed because any negative number can be expressed as the product of -1 and the opposite of the number.  -2 is the same as (-1)*2, for example.  Since {{{sqrt(ab)=sqrt(a)sqrt(b)}}}, if you had to deal with {{{sqrt(-2)}}}, you could change it to {{{sqrt(-1)sqrt(2)}}}.


What happens when the discriminant is negative, or less than 0, is this:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} where {{{D=b^2-4ac<0}}} becomes


{{{x = (-b +- sqrt( (-1)(-D) ))/(2*a) }}} but we say that {{{sqrt(-1)=i}}} so:


{{{x = (-b +- i*sqrt((-D) ))/(2*a) }}} (Note that if D < 0, -D > 0)


So D < 0 means that you will have a conjugate pair of complex roots of the form {{{alpha+-beta*i}}} where {{{alpha}}} and {{{beta}}} are real numbers and i is defined by {{{i^2=-1}}}  Furthermore, solutions to the general quadratic with D < 0 will be {{{alpha+-beta*i}}} where {{{alpha=-b/2a}}} and {{{beta=sqrt(-D)/2a}}}


Note that complex roots ALWAYS come in conjugate pairs.  If you find a complex root to a polynomial equation that is {{{alpha+beta*i}}}, then it is guaranteed that {{{alpha-beta*i}}} is also a root.  This is also true for higher degree polynomial equations.  It is no coincidence that quadratic (degree 2) equations have 2 roots.  In fact, cubics (3rd degree) equations have 3 roots, quartics (4th degree) equations have 4 roots, and nth degree equations have n roots.  As a consequence of the fact that complex roots must come in pairs, equations of odd degree (3rd, 5th, etc.) MUST have at least one real root.


Wait a minute, you say!  Yes, you have solved quadratics where you only got one answer.  In fact, you may have been told that if D=0 there is only 1 real root.  To that I say, 'not so fast, mathematics breath'.


If D=0 then you have a perfect square trinomial, something like {{{x^2-4x+4=0}}} which factors to {{{(x-2)(x-2)=0}}}.  [You can prove that by applying the quadratic equation to {{{x^2+2px+p^2}}} where {{{a=1}}}, {{{b=2p}}}, and {{{c=p^2}}}.  D will come out to be {{{4p^2-4p^2}}}]


Just because the factors are identical doesn't mean there aren't 2 of them.  And if there are 2 factors, there are also 2 roots.  They just happen to be identical.  Some mathematicians use the language 'one real root with a multiplicity of 2'  Same thing as far as I'm concerned.