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

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2 Answers

Let's consider the general second degree (quadratic) equation ax+ 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.
 
 

Comments

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