Roots of the Quadratic Equation

Date: 4/10/96 at 14:23:37
From: kelli rostkowski
Subject: Algebra: quadratic equations

Would you explain to me why a quadratic equation cannot have one 
irrational or one imaginary root.

Date: 4/27/96 at 19:42:54
From: Doctor Steven
Subject: Re: Algebra: quadratic equations

I'm assuming you mean a quadratic equation with rational 
coefficients. The reason why it can't have two complex roots is 
that if one complex number is a root, then its conjugate must be a 
root, in order to make the constant at the end of the quadratic a 
real number.  The reason it can't have just one irrational root is 
that an irrational multiplied by a rational is still irrational, 
so the constant at the end of the quadratic would be irrational.

To see this, say h and g are the roots and p(x) is the quadratic: 

   p(x) = x^2 + a*x + b

But p(x) is:

   (x - h)(x - g).

So (x - h)(x - g) = x^2 + a*x + b

Simplify the left side to get:

   x^2 + (-h - g)*x + h*g = x^2 + a*x + b.

So -h - g = a and h*g = b.

Both a and b must be rational.

If h is irrational and g isn't, then -h - g is irrational, and h*g 
must also be irrational.  But this can't happen, so one root can't 
be irrational and the other not.

If h is complex, and g is not its conjugate, then -h - g is still 
complex, and h*g is also complex then.  But this can't happen, so 
one root can't be complex and the other not its conjugate (which 
is also complex).

Hope this helps.

-Doctor Steven,  The Math Forum