Expert: Josh - 8/1/2005

Question
josh,
the section i am working on is entitled "the discriminant and equations that can be written in quadratic form". i know and understand the quadratic formula, i am also familiar with the concept that b^2-4ac is the format for the discriminate.  therefore x^2+x+1=0, the solution is -3.  where i am confused is my book states that since the solution is < 0 the solutions are two nonreal complex numbers that are complex conjugates.  i am not sure what they are talking about here.
thank-you in advance
james

Answer
Hi James,

The solution to the quadratic equation ax^2+bx+c=0 is given by x=[-b+sqrt(b^2-4ac)]/(2a) and [-b-sqrt(b^2-4ac)]/(2a). Letting the discriminant, D=b^2-4ac, we have
x=[-b+sqrt(D)]/(2a),[-b+sqrt(D)]/(2a).

The square root is only defined if the quantity D=b^2-4ac >=0. In your example, a=1,b=1,c=1. Since the "discriminant" (not the "solution"), D=-3, x=[-1+sqrt(-3)]/2, [-1-sqrt(-3)]/2. That is, the quadratic equation has no solution over the field of "real" numbers (the reason is because we cannot find a rational or irrational number that is the same as the square root of minus three).

However, over the field of complex numbers, the quadratic equation always has solutions that occur in complex conjugate pairs, like d+i*e and d-i*e, where "d" and "e" are real numbers and the imaginary number i is defined as i=sqrt(-1). Useful property to remember is that i*i=sqrt(-1)*sqrt(-1)=-1.

It follows that sqrt(-3)=sqrt(3*-1)=sqrt(3)*sqrt(-1)=sqrt(3)*i.
So, the solutions over the field of complex numbers may be expressed as x=-1/2+[sqrt(3)/2]*i and -1/2-[sqrt(3)/2]*i, respectively.

Some general interpretations on the quadratic solution are given in Q0008 in my question archive. You can have a look at the initial portion. The URL is www.geocities.com/joshcameron_ae

From a geometry perspective, all parabolas have a quadratic form.

[Once there, scroll down the page and click on the "Question Archive" link]

Cheers.