Let's consider the general second degree (quadratic) equation ax^{2 }+ bx + c = 0

If a quadratic expression on the left is to have factors, those factors will be of the form

(qx + p) and (sx + r) where p, q, r and s are all real numbers. So we now have the equation (qx + p) (sx + r) = 0

Either qx + p is 0 or sx + r is 0

or x = -p/q or -r/s which means x is rational. But we know from quadratic formula that x can be solved by

x = (-b + √(b^{2} - 4ac))/2a or (-b - √(b2 - 4ac))/2a

In order for x to be rational, clearly b^{2} - 4ac must be a perfect square

So from the above analysis, in order for a quadratic expression to have factors, b2 - 4ac must be a perfect square.

Galeboe M.

11/15/13