how can i see that a quadratic equation does not have factors evn before solving it

Question

This is a general question

 

Answer 1

If you have a quadratic equation of the form

AX²+BX+C = 0

 

The quadratic equation would return an imaginary number if 4*A*C > B²

Answer 2

Let's consider the general second degree (quadratic) equation ax² + 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 + √(b² - 4ac))/2a or (-b - √(b2 - 4ac))/2a

 

In order for x to be rational, clearly b² - 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.