What is true about the solutions of a quadratic equation when the radicand in the quadratic formula is negative?

1 Answer
May 18, 2015

When the radicand in the quadratic formula (the discriminant #Delta#) is negative it means that you cannot find pure Real solutions to your equation. This is not bad in the sense that you'll find Complex solutions.
These are solutions where appear the imaginary unit #i#.

Consider the following example:

The quadratic formula gave you:
#x_(1,2)=(6+-sqrt(-4))/2# we cannot do anything with the negative square root...or we can "camouflage" it a bit as:
#sqrt(-4)=sqrt(-1*4)=sqrt(-1)sqrt(4)=2sqrt(-1)=2i# where #i#, the imaginary unit, took the place of #sqrt(-1)#.
Ok it is kind of cheating but it works!

You can now overcome the problem of the negative square root and write the two solutions as:

#x_(1,2)=(6+-2i)/2=3+-i# which are two complex conjugate numbers. The good thing is that if you substitute back these two "strange" solutions in your original equations they work (remembering the meaning of #i#).

Hope it helps!