Question 41299
By D you meant determinant. Right?


For a quadratic equation with real coefficients, if D < 0 then the roots are imaginery and occur in conjugate pairs as well.
Thus if {{{ax^2 + bx + c = 0}}} be a quadratic equation where a,b,c are real then if {{{b^2-4*a*c <0}}} then the roots are of the form A + iB and A - iB where A and B are real.


Example: {{{5x^2 + 8x + 5 = 0}}} is a quadratic equation with real coefficients.
The determinant is {{{8^2 - 4*5*5}}} = 64 - 100 = -36 < 0.
Now let us find the roots.
Sreedharachary's formula is
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
or {{{x = (-8 +- sqrt( -36 ))/10 }}}
or {{{x = (-8 +- sqrt( (6*i)^2 ))/10 }}}
or {{{x = (-8 +- 6*i)/10 }}} = {{{-0.8 +- 0.6*i }}}
Thus either x = -0.8 + 0.6i or x = -0.8 - 0.6i


So you can see that the roots are complex conjugate.