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

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Imtiazur S. | Tutor with master's degree and experience teaching CalculusTutor with master's degree and experienc...
4.9 4.9 (336 lesson ratings) (336)
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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.


Richard L. | Engineer Tutor - Physics, Calculus, and Pre-CalculusEngineer Tutor - Physics, Calculus, and ...
4.7 4.7 (3 lesson ratings) (3)
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